Trigonometric Ratios of Multiple Angles

Trigonometric ratios of Multiple Angles

example for Trigonometric ratios of Multiple Angles
example for Trigonometric ratios of Multiple Angles

If A is an angle, then 2A, 3A, 4A, 5A, etc. are called multiple angles of A. In this section we will discuss about the trigonometric ratios of angles 2A and 3A in terms of A.

(a) sin2A = sin(A + A) = sinA. cosA + cosA. sinA = 2sinA. cosA

(b) sin2A = 2sinA. cosA = \(\frac{2sinA. cosA}{1}\) =\(\frac{2sinA. cosA}{cos^{2}A + sin^{2}A}\) =\(\frac{\frac{2sinA. cosA}{cos^2A}}{\frac{cos^2A}{cos^2A} + \frac{sin^2A}{cos^2A}}\) = \(\frac{2tanA}{1 + tan^2A}\)

(c) sin2A =\(\frac{2tanA}{1 + tan^2 A}\) =\(\frac{\frac{2}{cotA}}{1 +\frac{1}{cot^2 A}}\) =\(\frac{2}{cotA}\)x\(\frac{cot^2A}{1 + cot^2 A}\) =\(\frac{2cotA}{1 + cot^2 A}\)

(d) cos2A = cos(A + A) = cosA. cosA - sinA. sinA = cos2A - sin2A

(e) cos2A = cos2A - sin2A = 1 - sin2A - sin2A = 1 - 2sin2A

(f) cos2A = cos2A - sin2A = cos2A - (1 - cos2A) = 2cos2A - 1

(g) cos2A = cos2A - sin2A =\(\frac{cos^2A - sin^2A}{1}\) =\(\frac{cos^2A - sin^2A}{cos^2A + sin^2A}\) =\(\frac{\frac{cos^2A}{cos^2A} - \frac{sin^2A}{cos^2A}}{\frac{cos^2A}{cos^2A} + \frac{sin^2A}{cos^2A}}\) =\(\frac{1 - tan^2A}{1 + tan^2A}\)

(h) cos2A =\(\frac{1 - tan^2A}{1 + tan^2A}\) =\(\frac{1 - \frac{1}{cot^2A}}{1 +\frac{1}{cot^2A}}\) =\(\frac{cot^2A - 1}{cot^2A + 1}\)

(i) tan2A = tan(A + B) =\(\frac{tanA + tanA}{1 - tanA. tanA}\) =\(\frac{2tanA}{1 - tan^2A}\)

(j) tan2A =\(\frac{2tanA}{1 - tan^2A}\) =\(\frac{\frac{2}{cotA}}{1 -\frac{1}{cot^2A}}\) =\(\frac{2}{cotA}\) x \(\frac{cot^2A}{cot^2A - 1}\)

(k) cot2A = cot (A + A) =\(\frac{cotA. cotA - 1}{cotA + cotA}\) =\(\frac{cot^2A - 1}{2cotA}\)

(l) cot2A =\(\frac{cot^2A - 1}{2cotA}\) =\(\frac{\frac{1}{tan^2A} - 1}{\frac{2}{tanA}}\) =\(\frac{1 - tan^2A}{tan^2A}\) x \(\frac{1 - tan^2 A}{2 tanA}\) =\(\frac{1 - tan^2 A}{2 tanA}\)

Some useful results

(a) 1 + cos2A = 1 + cos2A - sin2A = 1 - sin2A + cos2A = cos2A + cos2A = 2cos2A

(b) 1 - cos2A = 1 - (cos2A - sin2A) = 1 - cos2A + sin2A = sin2A + sin2A = 2sin2A

(c) 1 + sin2A = cos2A + sin2A + 2 sinA .cosA = cos2A + 2 cosA . sinA + sin2A

= (cosA + sinA)2

(d) 1 - sin2A = cos2A + sin2A - 2sinA .cosA = cos2A - 2 cosA .sinA + sin2A

= (cosA - sinA)2

Trigonometric ratios of 3A in terms of A

(a) sin3A = sin(2A + A) = sin2A. cosA + cos2A. sinA

= 2 sinA . cosA. cosA + (1 - 2sin2A). sinA

= 2 sinA (1 - sin2A) + sinA - 2sin\(^3\)A = 2sinA. 2sin\(^3\)A + sinA - 2 sin\(^3\)A = 3 sinA - 4 sin\(^3\)A

(b) cos3A = cos (2a + A) = cos2A. cosA - sin2A. sinA

= (2cos2A - 1) cosA - 2sinA. cosA. sinA = 2cos\(^3\)A - cosA - 2cosA (1 - cos2A)

= 2 cos\(^3\)A - cosA - 2cosA + 2cos\(^3\)A = 4 cos\(^3\)A - 3 cosA.

(c) tan3A = tan( 2A + A) =\(\frac{tan2A + tanA}{1 - tan2A. tanA}\) =\(\frac{\frac{2 tan A}{1 - tan^2 A} + tanA}{1 -\frac{2tan A}{1 - tan^2 A}. tanA}\)

=\(\frac{2 tan A + tan A - tan^3 A}{1 - tan^2 A - 2 tan^2 A}\) =\(\frac{3 tan A - tan^3 A}{1 - 3 tan^2 A}\)

Geometrical proof of 2A formulae :

.

Let O be the centre of the circle ABC and AB be a diameter. Let ∠CAB = A.

Then ∠COB = 2A

Here, ∠ACB = 900. Let CM is perpendicular to AB.

Then ∠ACM = 900 - A and hence. ∠BCM = A.

Now. sin2A =\(\frac{CM}{OC}\) =\(\frac{2CM}{2OC}\) =\(\frac{2CM}{AB}\)

= 2 \(\frac{CM}{AC}\). \(\frac{AC}{AB}\) = 2sinA. cosA

cos2A =\(\frac{OM}{OC}\) =\(\frac{2OM}{2OC}\) =\(\frac{2OM}{AB}\)

= \(\frac{(AO + OM) - (AO - OM)}{AB}\) =\(\frac{AM - BM}{AB}\)

= \(\frac{AM}{AB}\) - \(\frac{BM}{AB}\) = \(\frac{AM}{AC}\). \(\frac{AC}{AB}\) - \(\frac{BM}{BC}\). \(\frac{BC}{AB}\)

= cosA. cosA - sinA. sinA = cos2A - sin2A

tan2A = \(\frac{CM}{OM}\) = \(\frac{2CM}{2OM}\) =\(\frac{2CM}{(AO + OM) - (AO -OM)}\) = \(\frac{2CM}{(AO + OM) - (BO - OM)}\) =\(\frac{2CM}{AM - BM}\)

=\(\frac{\frac{2CM}{AM}}{\frac{AM}{AM} - \frac{BM}{AM}}\) =\(\frac{\frac{2CM}{AM}}{1 -\frac{BM}{CM}× \frac{CM}{AM}}\)

=\(\frac{2 tan A}{1 - tan A. tan A}\)

=\(\frac{2 tan A}{1 - tan^2 A}\)

Multiple angles formulae

sin2A = 2sinA. cosA tan2A =\(\frac{2cot A}{cot^2 A - 1}\)
sin2A =\(\frac{2tan A}{1 + tan^2 A}\) cot2A =\(\frac{cot^2 A - 1}{2cot A}\)
sin2A =\(\frac{2cot A}{1 + cot^2 A}\) cot2A =\(\frac{1 - tan^2 A}{2tan A}\)
cos2A = cos2A - sin2A 1 + cos2A = 2cos2A
cos2A = 1 - 2sin2A 1 - cos2A = 2 sin2A
cos2A = 2cos2A - 1 1 + sin2A = (cosA + sinA)2
cos2A =\(\frac{1 - tan^2 A}{1 + ttan^2 A}\) 1 - sin2A = (cosA - sinA)2
cos2A =\(\frac{cot^2 A - 1}{cot^2 A + 1}\) sin3A = 3 sinA - 4 sin \(^3\) A
tan2A =\(\frac{2 tan A}{1 - tan^2 A}\) cos3A = 4 cos\(^3\)A - 3cos A

tan3 A =\(\frac{3 tan A - tan^3 A}{1 - 3tan^2 A}\)

sin2A = 2sinA. cosA tan2A =\(\frac{2cot A}{cot^2 A - 1}\)
sin2A =\(\frac{2tan A}{1 + tan^2 A}\) cot2A =\(\frac{cot^2 A - 1}{2cot A}\)
sin2A =\(\frac{2cot A}{1 + cot^2 A}\) cot2A =\(\frac{1 - tan^2 A}{2tan A}\)
cos2A = cos2A - sin2A 1 + cos2A = 2cos2A
cos2A = 1 - 2sin2A 1 - cos2A = 2 sin2A
cos2A = 2cos2A - 1 1 + sin2A = (cosA + sinA)2
cos2A =\(\frac{1 - tan^2 A}{1 + ttan^2 A}\) 1 - sin2A = (cosA - sinA)2
cos2A =\(\frac{cot^2 A - 1}{cot^2 A + 1}\) sin3A = 3 sinA - 4 sin \(^3\) A
tan2A =\(\frac{2 tan A}{1 - tan^2 A}\) cos3A = 4 cos\(^3\)A - 3cos A

tan3 A =\(\frac{3 tan A - tan^3 A}{1 - 3tan^2 A}\)

L.H.S.

=\(\frac {sin\theta + sin2\theta}{1 + cos\theta + cos2\theta}\)

= \(\frac {sin\theta + 2sin\theta cos\theta}{1 + cos\theta + 2cos^2\theta - 1}\)

= \(\frac {sin\theta (1 + 2cos\theta)}{cos\theta (1 + 2cos\theta)}\)

= \(\frac {sin\theta}{cos\theta}\)

= tan\(\theta\)

Hence, L.H.S. = R.H.S. proved

Here,

sin3A cos2A

= (3 sinA - 4 sin3A) (1 - sin2A)

= 3 sinA - 4 sin3A - 3 sin3A + 4 sin5A

= 4 sin5A - 7 sin3A + 3 sinA Ans

Here,

L.H.S.

= cos 2A

= \(\frac {cos^2A - sin^2A}{1}\)

= \(\frac {cos^2A - sin^2A}{cos^2A + sin^2A}\)

=\(\frac {{\frac {cos^2A}{sin^2A}} - {\frac {sin^2A}{cos^2A}}}{{\frac {cos^2A}{cos^2A}} + {\frac {sin^2A}{cos^2A}}}\)

= \(\frac {1 - tan^2A}{1 + tan^2A}\)

Hence, L.H.S. = R.H.S. Proved

Here,

tan A = \(\frac 34\)

sin 2A

= \(\frac {2tan A}{1 + tan^2A}\)

= \(\frac {2 × \frac 34}{1 + (\frac {3}{4})^2}\)

= \(\frac {\frac 32}{1 + \frac {9}{16}}\)

= \(\frac 32\)× \(\frac {16}{25}\)

= \(\frac {24}{25}\) Ans

cos 2A

= \(\frac {1 - tan^2A}{1 + tan^2A}\)

= \(\frac{1 -(\frac {3}{4})^2}{1 + (\frac {3}{4})^2}\)

= \(\frac {1 - \frac 9{16}}{1 +\frac 9{16}}\)

= \(\frac {\frac {16 - 9}{16}}{\frac {16 + 9}{16}}\)

= \(\frac {7}{16}\)× \(\frac {16}{25}\)

= \(\frac 7{25}\) Ans

L.H.S.

=\(\frac {sin 2A}{1 + cos 2A}\)

= \(\frac {2 sin A cos A}{2 cos^2A}\)

= \(\frac {sin A}{cos A}\)

= tan A

Hence, L.H.S. = R.H.S. Proved

Here,

cos\(\alpha\) = \(\frac {\sqrt 3}{2}\)

sin\(\alpha\)

= \(\sqrt {1 - cos^2\alpha}\)

= \(\sqrt {1 - (\frac {\sqrt 3}{2})^2}\)

=\(\sqrt {1 - \frac 34}\)

=\(\sqrt {\frac {4 - 3}{4}}\)

=\(\sqrt {\frac 14}\)

= \(\frac 12\)

sin3\(\alpha\)

= 3 sin\(\alpha\) - 4 sin3\(\alpha\)

= 3× \(\frac 12\) - 4× (\(\frac 12\))3

= \(\frac 32\) - \(\frac 48\)

= \(\frac 32\) - \(\frac 12\)

= \(\frac {3 - 1}{2}\)

= \(\frac 22\)

= 1 Ans

Again,

cos 3\(\alpha\)

= 4 cos2\(\alpha\) - 3 cos\(\alpha\)

= 4 (\(\frac {\sqrt 3}{2}\))3- 3× \(\frac {\sqrt 3}{2}\)

= \(\frac {12\sqrt 3}{8}\) - \(\frac {3\sqrt 3}{2}\)

= \(\frac {3\sqrt 3}{2}\) - \(\frac {3\sqrt 3}{2}\)

= 0 Ans

L.H.S.

=4 (cos310° + sin320°)

=4cos310° + 4sin320°

= [cos 3(10°) + 3 cos 10°] + [3 sin20° - sin 3(20°)]

= cos 30° + 3 cos 10° + 3 sin 20° - sin 60°

= \(\frac {\sqrt 3}{2}\) + 3 (cos 10° + sin 20°) - \(\frac {\sqrt 3}{2}\)

=3 (cos 10° + sin 20°)

Hence, L.H.S. = R.H.S. Proved

Here,

cos\(\theta\) = \(\frac {12}{13}\) and sin\(\alpha\) = \(\frac 45\)

sin\(\theta\)

= \(\sqrt {1 - cos^2\theta}\)

= \(\sqrt {1 - (\frac {12}{13})^2}\)

= \(\sqrt {\frac {169 - 144}{169}}\)

= \(\sqrt {\frac {25}{169}}\)

= \(\frac 5{13}\)

cos\(\alpha\)

= \(\sqrt {1 - sin^2\alpha}\)

= \(\sqrt {1 - (\frac {4}{5})^2}\)

= \(\sqrt {\frac {25 - 16}{25}}\)

= \(\sqrt {\frac 9{25}}\)

= \(\frac 35\)

∴ sin 2\(\theta\) = 2 sin\(\theta\) cos\(\theta\) = 2× \(\frac 5{13}\)× \(\frac {12}{13}\) = \(\frac {120}{169}\) Ans

∴ cos 2\(\alpha\) = 2 cos2\(\alpha\) - 1 = 2× \(\frac 35\)× \(\frac 35\) - 1 = \(\frac {18 - 25}{25}\) = -\(\frac 7{25}\) Ans

Here,

3\(\theta\) + \(\theta\) = 4\(\theta\)

Putting tan on both,

tan (3\(\theta\) + \(\theta\)) = tan 4\(\theta\)

or, \(\frac {tan 3\theta + tan\theta}{1 - tan 3\theta tan\theta}\) = tan 4\(\theta\)

or, tan 3\(\theta\) + tan\(\theta\) = tan 4\(\theta\) - tan\(\theta\) tan 3\(\theta\) tan 4\(\theta\)

or,tan\(\theta\) tan 3\(\theta\) tan 4\(\theta\) =tan 4\(\theta\) -tan 3\(\theta\) -tan\(\theta\)

Hence, L.H.S. = R.H.S. proved

L.H.S.

=\(\frac {sin 5\theta}{sin \theta}\) - \(\frac {cos 5\theta}{cos \theta}\)

= \(\frac {sin 5\theta cos \theta - cos 5\theta sin \theta}{sin \theta cos \theta}\)

= \(\frac {sin (5\theta - \theta)}{sin\theta cos\theta}\)

= \(\frac {sin 4\theta}{sin\theta cos\theta}\)

= \(\frac {sin 2(2\theta)}{sin\theta cos\theta}\)

= \(\frac {2 sin 2\theta cos 2\theta}{sin\theta cos\theta}\)

= \(\frac {2 × 2 sin\theta cos\theta cos2\theta}{sin\theta cos\theta}\)

= 4 cos 2\(\theta\)

Hence, L.H.S. = R.H.S. Proved

L.H.S.

= \(\frac {cos^3A + sin^3A}{cosA + sinA}\)

= \(\frac {(cosA + sinA)(cos^2A + sin^2A - sinA cosA)}{cosA + sinA}\)

= 1 - \(\frac {2 sinA cosA}{2}\)

= 1 - \(\frac 12\)sin 2A

Hence, L.H.S. = R.H.S. Proved

L.H.S.

=\(\frac 1{tan3A - tanA}\) - \(\frac 1{cot3A - cotA}\)

=\(\frac 1{tan3A - tanA}\) - \(\frac 1{{\frac 1{tan3A}} - {\frac 1{tanA}}}\)

=\(\frac 1{tan3A - tanA}\) - \(\frac {\frac 1{tanA - tan3A}}{tan3A tanA}\)

=\(\frac 1{tan3A - tanA}\) - \(\frac {tan 3A tanA}{tanA - tan3A}\)

= \(\frac 1{tan3A - tanA}\) +\(\frac {tan 3A tanA}{tan3A - tanA}\)

= \(\frac {1 + tan 3A tanA}{tan3A - tanA}\)

= \(\frac {\frac 1{tan3A - tanA}}{1 + tan3A tanA}\)

= \(\frac 1{cot (3A - A)}\)

= \(\frac 1{cot2A}\)

= tan 2A

Hence, L.H.S. = R.H.S. proved

L.H.S.

=\(\frac {cos\theta}{cos\theta - sin\theta}\) -\(\frac {cos\theta}{cos\theta +sin\theta}\)

= \(\frac {cos\theta (cos\theta + sin\theta) - cos\theta (cos\theta - sin\theta)}{(cos\theta - sin\theta) (cos\theta + sin\theta)}\)

= \(\frac {cos^2\theta + cos\theta sin\theta - cos^2\theta + cos\theta sin\theta}{cos^2\theta - sin^2\theta}\)

= \(\frac {2 sin\theta cos\theta}{cos^2\theta - sin^2\theta}\)

= \(\frac {sin 2\theta}{cos 2\theta}\)

= tan 2\(\theta\)

Hence, L.H.S. = R.H.S. Proved

L.H.S.

=\(\frac {1 - cos 2A + sin2A}{1 + cos 2A + sin2A}\)

= \(\frac {2sin^2A + 2sinAcosA}{2cos^2A + 2sinAcosA}\) [\(\because\) 1 - cos2A = 2sin2A and 1 + cos2A = 2cos2A]

= \(\frac {2sin A (sinA + cosA)}{2cosA (cosA + sinA)}\)

= tanA

Hence, L.H.S. = R.H.S. proved

L.H.S.

=\(\frac {1 - tan^2(\frac {π}{4} - \theta)}{1 + tan^2(\frac {π}{4} - \theta)}\)

= cos 2(\(\frac {π}{4} - \theta)\)

= cos(\(\frac {π}{2} - 2\theta)\)

= sin 2\(\theta\)

hence, L.H.S. = R.H.S. Proved

L.H.S.

=\(\frac {sin 3A}{sinA}\) - \(\frac {cos 3A}{cosA}\)

= \(\frac {3 sinA - 4 sin^3A}{sinA}\) - \(\frac {4 cos^3A - 3 cosA}{cosA}\)

= \(\frac {sinA (3 - 4 sin^2A)}{sinA}\) - \(\frac {cosA (4 cos^2A - 3)}{cosA}\)

= 3 - 4 (1 - cos2A) - 4 cos2A + 3

= 3 - 4 + 4 cos2A - 4 cos2A + 3

= 6 - 4

= 2

Hence, L.H.S = R.H.S. Proved

L.H.S.

=4 cosec 2A . cot 2A

= 4(\(\frac 1{sin2A})\) . \(\frac {cos 2A}{sin 2A}\)

= \(\frac {4 (cos^2A - sin^2A)}{2 sinA cosA . 2 sinA cosA}\)

= \(\frac {cos^2A - sin^2A}{sin^2A cos^2A}\)

= \(\frac {cos^2A}{sin^2A cos^2A}\) - \(\frac {sin^2A}{sin^2A cos^2A}\)

= \(\frac {1}{sin^2A}\) - \(\frac {1}{cos^2A}\)

= cosec2A - sec2A

Hence, L.H.S. = R.H.S. proved

Here,

cosA = \(\frac {15}{17}\)

cosA

= \(\sqrt {1 - cos^2A}\)

= \(\sqrt {1 - (\frac {15}{17}})^2\)

= \(\sqrt {1 - \frac {225}{289}}\)

= \(\sqrt {\frac {289 - 225}{289}}\)

= \(\sqrt {\frac {64}{289}}\)

= \(\frac 8{17}\)

∴ tanA = \(\frac {sinA}{cosA}\) = \(\frac {\frac {8}{17}}{\frac {15}{17}}\) = \(\frac 8{15}\)

Now,

tan 3A

= \(\frac {3 tanA - tan^3A}{1 - 3 tan^2A}\)

= \(\frac {3 × \frac 8{15} - (\frac 8{15})^3}{1 - 3 ×(\frac 8{15})^2}\)

= \(\frac {\frac {24}{15} - \frac {512}{3375}}{1 - \frac {192}{225}}\)

= \(\frac {\frac {5400 - 512}{3375}}{\frac {225 - 192}{225}}\)

= \(\frac {\frac {4888}{3375}}{\frac {33}{225}}\)

= \(\frac {4888}{3375}\)× \(\frac {225}{33}\)

= \(\frac {4888}{15 × 33}\)

= \(\frac {4888}{495}\) Ans

L.H.S.

=\(\frac 1{tan 3A + tanA}\) - \(\frac 1{cot 3A + cotA}\)

= \(\frac 1{tan 3A + tanA}\) - \(\frac 1{{\frac 1{tan 3A}} + {\frac 1{tanA}}}\)

= \(\frac 1{tan 3A + tanA}\) - \(\frac {\frac 1{tanA + tan3A}}{tan3A tanA}\)

= \(\frac 1{tan 3A + tanA}\) - \(\frac {tan 3A tanA}{tan 3A + tanA}\)

= \(\frac {1 - tan 3A tanA}{tan 3A + tanA}\)

= \(\frac {\frac 1{tan 3A + tanA}}{1 - tan3A tanA}\)

= \(\frac 1{tan (3A + A)}\)

= \(\frac 1{tan 4A}\)

= cot 4A

Hence, L.H.S. = R.H.S. Proved

L.H.S.

=\(\frac {1 + sin 2A}{cos 2A}\)

= \(\frac {sin^2A + cos^2A + 2 sinA cosA}{cos^2A - sin^2A}\)

= \(\frac {(cosA + sinA)^2}{(cosA + sinA) (cosA - sinA)}\)

= \(\frac {cosA + sinA}{cosA - sinA}\)

Hence, L.H.S. = R.H.S. Proved

Here,

sin 2\(\theta\)

= \(\frac {2 tan\theta}{1 + tan^2\theta}\)

= \(\frac {2 × \frac 43}{1 + (\frac 43)^2}\)

= \(\frac {\frac 83}{1 + \frac {16}{9}}\)

= \(\frac {\frac83}{\frac {9 + 16}{9}}\)

= \(\frac {\frac 83}{\frac {25}9}\)

= \(\frac 83\)× \(\frac 9{25}\)

= \(\frac {24}{25}\) Ans

Here,

cos A = \(\frac 35\)

sinA

= \(\sqrt {1 - cos^2 A}\)

= \(\sqrt {1 - (\frac 35)^2}\)

= \(\sqrt {1 - \frac 9{25}}\)

= \(\sqrt {\frac {25 - 9}{25}}\)

= \(\sqrt {\frac {16}{25}}\)

= \(\frac 45\)

Now,

sin 2A

= 2 sinA cosA

= 2× \(\frac 45\)× \(\frac 35\)

= \(\frac {24}{25}\) Ans

L.H.S.

= sin4A

= (sin2A)2

=[\(\frac 12\) (1 - cos 2A)]2

= \(\frac 14\) [1 - 2 cos 2A + cos2A]

= \(\frac14\) [1 - 2 cos2A + (\(\frac {1 + cos4A}{2}\))]

= \(\frac 14\) [\(\frac {2 - 4 cos2A + 1 + cos4A}2\)]

= \(\frac 18\)(3 - 4 cos 2A + cos 4A)

Hence, L.H.S. = R.H.S. Proved

Here,

cosec 2A + cot 4A = cotA - cosec 4A

or, cot 4A + cosec 4A = cotA - cosec 2A

L.H.S.

= cot 4A + cosec 4A

= \(\frac {cos 4A}{sin 4A}\) + \(\frac 1{sin 4A}\)

= \(\frac {cos 4A + 1}{sin 4A}\)

= \(\frac {2 cos^22A - 1 + 1}{2 sin 2A cos 2A}\)

= \(\frac {2 cos^22A}{2 sin 2A cos 2A}\)

= \(\frac {cos 2A}{sin 2A}\)

= \(\frac {2 cos^2A - 1}{2 sinA cosA}\)

= \(\frac {2 cos^2A}{2 sinA cosA}\) - \(\frac 1{2 sinA cosA}\)

= cotA - \(\frac 1{sin 2A}\)

= cotA - cosec 2A

Hence, L.H.S. = R.H.S. Proved

L.H.S.

=\(\frac {\sqrt 3}{sin 20°}\) - \(\frac 1{cos 20°}\)

= \(\frac {\sqrt 3 cos 20° - sin 20°}{sin 20° cos 20°}\)

= \(\frac {2 (\frac {\sqrt 3}{2} cos 20° - \frac 12 sin 20°)}{sin 20° cos 20°}\)

= \(\frac {2 × 2 (sin 60° cos 20° - cos 60° sin 20°)}{2 sin 20° cos 20°}\)

= \(\frac {4 [sin (60° - 20°)]}{sin 2 × 20°}\)

= \(\frac {4 sin 40°}{sin 40°}\)

= 4

Hence, L.H.S. = R.H.S. Proved

L.H.S.

=\(\frac {1}{sin 10°}\) - \(\frac {\sqrt 3}{cos 10°}\)

= \(\frac {cos 10° - \sqrt 3 sin 10°}{sin 10° cos 10°}\)

= \(\frac {2 (\frac {1}{2} cos 10° - \frac {\sqrt 3}2 sin 10°)}{sin 10° cos 10°}\)

= \(\frac {2 × 2 (sin 30° cos10° - cos 30° sin 10°)}{2 sin 10° cos 10°}\)

= \(\frac {4 [sin (30° - 10°)]}{sin 2 × 10°}\)

= \(\frac {4 sin 20°}{sin 20°}\)

= 4

Hence, L.H.S. = R.H.S. Proved

L.H.S.

= cos 5A

= cos (2A + 3A)

= cos 2A cos 3A - sin 2A sin 3A

= (2 cos2A - 1) (4 cos3A - 3 cosA) - 2 sinA cosA (3 sinA - 4 sin3A)

= 8 cos5A - 6 cos3A - 4 cos3A + 3 cosA - 6 sin2A cosA + 8 sin4A cosA

= 8 cos5A - 10 cos3A + 3 cosA - 6 (1 - cos2A) cosA + 8 (1 - cos2A)2 cosA

= 8 cos5A - 10 cos3A + 3 cosA - 6 cosA + 6 cos3A + 8 cosA (1 - 2 cos2A + cos4A)

= 8 cos5A - 4 cos3A - 3 cosA + 8 cosA - 16 cos3A + 8 cos5A

=16 cos5A - 20 cos3A + 5 cosA

Hence, L.H.S. = R.H.S. Proved

L.H.S.

=tan (\(\frac {\pi}{4} + A\)) + tan(\(\frac {\pi}{4} -A\))

= \(\frac {tan \frac {\pi}{4} + tanA}{1 - tan {\frac {\pi}{4}} tanA}\) +\(\frac {tan \frac {\pi}{4} -tanA}{1 +tan {\frac {\pi}{4}} tanA}\)

= \(\frac {1 + tanA}{1 - 1 × tanA}\) + \(\frac {1 -tanA}{1 +1 × tanA}\)

= \(\frac {(1 + tanA)^2 + (1 - tan A)^2}{(1 - tanA) (1 + tanA)}\)

= \(\frac {1 + 2 tanA + tan^2A + 1 - 2 tanA + tan^2A}{1 - tan^2A}\)

= \(\frac {2 + 2 tan^2A}{1 - tan^A}\)

= \(\frac {2 (1 + tan^2A)}{1 - tan^2A}\)

= \(\frac 2{\frac {1 - tan^2A}{1 + tan^2A}}\)

= \(\frac 2{cos 2A}\)

= 2 sec 2A

Hence, L.H.S. = R.H.S. Proved

Here,

tan\(\theta\) = \(\frac 56\) and tan\(\beta\) = \(\frac 1{11}\)

tan (\(\theta\) + \(\beta\)) = \(\frac {tan \theta + tan \beta}{1 - tan\theta tan\beta}\)

or,tan (\(\theta\) + \(\beta\)) = \(\frac {\frac 56 + \frac 1{11}}{1 - \frac 56 × \frac 1{11}}\)

or, tan (\(\theta\) + \(\beta\)) = \(\frac {\frac {5 × 11 + 1 × 6}{66}}{\frac {66 - 5}{66}}\)

or,tan (\(\theta\) + \(\beta\)) = \(\frac {55 + 6}{66}\)× \(\frac {66}{61}\)

or,tan (\(\theta\) + \(\beta\)) = \(\frac {61}{66}\)× \(\frac {66}{61}\)

or,tan (\(\theta\) + \(\beta\)) = 1

or,tan (\(\theta\) + \(\beta\)) = tan 45°

or,tan (\(\theta\) + \(\beta\)) = tan\(\frac {\pi^c}{4}\)

∴ (\(\theta\) + \(\beta\)) =\(\frac {\pi^c}{4}\)

Hence, L.H.S. = R.H.S. Proved

L.H.S.

=cos6\(\theta\) + sin6\(\theta\)

= (cos2\(\theta\))3 + (sin2\(\theta\))3

= (cos2\(\theta\) + sin2\(\theta\)) (cos4\(\theta\) - cos2\(\theta\) sin2\(\theta\) + sin4\(\theta\))

= {(cos2\(\theta\))2 - cos2\(\theta\) sin2\(\theta\) + (sin2\(\theta\))2}

= {(cos2\(\theta\) + sin2\(\theta\))2 - 2 cos2\(\theta\) sin2\(\theta\) - cos2\(\theta\) sin2\(\theta\)}

= 1 - 3 sin2\(\theta\) cos2\(\theta\)

= 1 - 3 . \(\frac 44\) (sin2\(\theta\) cos2\(\theta\))

= 1 - \(\frac 34\) (4 sin2\(\theta\) cos2\(\theta\))

= 1 - \(\frac 34\) (2 sin\(\theta\) cos\(\theta\))2

= 1 - \(\frac 34\) (sin 2\(\theta\))2

= 1 - \(\frac 34\) (sin22\(\theta\))

= 1 - \(\frac 34\) (1 - cos22\(\theta\))

= \(\frac {4 - 3 + 3 cos^22\theta}{4}\)

= \(\frac {1 + 3 cos^22\theta}{4}\)

= \(\frac 14\) (1 + 3 cos22\(\theta\))

Hence, L.H.S. = R.H.S. Proved

L.H.S.

=\(\frac {2 cos 8\theta + 1}{2 cos\theta + 1}\)

=\(\frac {2 cos 2 × 4\theta + 1}{2 cos\theta + 1}\)

=\(\frac {2 (2cos^2 4\theta - 1) + 1}{2cos\theta + 1}\)

=\(\frac {4 cos^2 4\theta -2 + 1}{2cos\theta + 1}\)

=\(\frac {4 cos^2 4\theta - 1}{2cos\theta + 1}\)

=\(\frac {(2cos 4\theta)^2 - (1)^2}{2cos\theta + 1}\)

=\(\frac {(2 cos 4\theta + 1)(2 cos 4\theta - 1)}{2cos\theta + 1}\)

=\(\frac {(2 cos 2 × 2\theta + 1)(2 cos 4\theta - 1)}{2cos\theta + 1}\)

=\(\frac {[2 (2cos^2\theta - 1) + 1](2 cos 4\theta - 1)}{2cos\theta + 1}\)

=\(\frac {(4 cos^2\theta - 2 + 1)(2 cos 4\theta - 1)}{2cos\theta + 1}\)

=\(\frac {(4 cos^2\theta - 1)(2 cos 4\theta - 1)}{2cos\theta + 1}\)

=\(\frac {[(2 cos2\theta)^2 - (1)^2](2 cos 4\theta - 1)}{2cos\theta + 1}\)

=\(\frac {(2 cos 2\theta + 1) (2 cos 2\theta - 1) (2 cos 4\theta - 1)}{2cos\theta + 1}\)

=\(\frac {[2 × (2 cos^2\theta - 1) + 1] (2 cos 2\theta - 1) (2 cos 4\theta - 1)}{2cos\theta + 1}\)

=\(\frac {(4 cos^2\theta - 2 + 1) (2 cos 2\theta - 1) (2 cos 4\theta - 1)}{2cos\theta + 1}\)

=\(\frac {(4 cos^2\theta - 1) (2 cos 2\theta - 1) (2 cos 4\theta - 1)}{2cos\theta + 1}\)

=\(\frac {[(2 cos\theta)^2 - (1)^2] (2 cos 2\theta - 1) (2 cos 4\theta - 1)}{2cos\theta + 1}\)

=\(\frac {(2 cos\theta + 1) (2 cos \theta - 1) (2 cos 2\theta - 1) (2 cos 4\theta - 1)}{2cos\theta + 1}\)

=(2 cos\(\theta\) - 1) (2 cos 2\(\theta\) - 1) (2 cos 4\(\theta\) - 1)

Hence, L.H.S. = R.H.S. Proved

L.H.S.

= cos2A + sin2A cos 2B

= cos2A + sin2A(1 - 2 sin2B)

= cos2A + sin2A - 2 sin2A sin2B

= 1 - 2 sin2A sin2B

= 1 - (1 - cos 2A) sin2B

= 1 - sin2B + sin2B cos 2A

= cos2B + sin2B cos 2A

Hence, L.H.S. = R.H.S. Proved

Here,

cos 4\(\theta\)

= cos 2 (2\(\theta\))

= 1 - 2 sin2 (2\(\theta\))

= 1 - 2 (2sin\(\theta\) cos\(\theta\))2

= 1 - 2× 4 sin2\(\theta\) cos2\(\theta\)

= 1 - 8 sin2\(\theta\) (1 - sin2\(\theta\))

= 1 - 8 sin2\(\theta\) + 8 sin4\(\theta\) Ans

Here,

tan2\(\alpha\) = 1 + 2 tan2\(\beta\)

or, \(\frac {1 - cos 2\alpha}{1 + cos 2\alpha}\) = 1 + 2 (\(\frac {1 - cos 2\beta}{1 + cos 2\beta})\)

or,\(\frac {1 - cos 2\alpha}{1 + cos 2\alpha}\) = \(\frac {1 + cos 2\beta + 2 - 2 cos 2\beta}{1 + cos 2\beta}\)

or,\(\frac {1 - cos 2\alpha}{1 + cos 2\alpha}\) = \(\frac {3 - cos 2\beta}{1 + cos 2\beta}\)

or,\(\frac {1 - cos 2\alpha}{1 + cos 2\alpha}\) + 1 = \(\frac {3 - cos 2\beta}{1 + cos 2\beta}\) + 1

or, \(\frac {1 - cos 2\alpha + 1 + cos 2\alpha}{1 + cos 2\alpha}\) = \(\frac {3 - cos 2\beta + 1 + cos 2\beta}{1 + cos 2\beta}\)

or, \(\frac 2{1 + cos 2\alpha}\) = \(\frac 4{1 + cos 2\beta}\)

or, 2 (1 + cos 2\(\beta\)) = 4 (1 + cos 2\(\alpha\))

or, 2 + 2 cos 2\(\beta\) = 4 + 4 cos 2\(\alpha\)

or, 2 cos 2\(\beta\) =4 + 4 cos 2\(\alpha\) - 2

or,2 cos 2\(\beta\) = 2+ 4 cos 2\(\alpha\)

or, cos 2\(\beta\) = \(\frac {2(1 + 2 cos 2\alpha)}{2}\)

or,cos 2\(\beta\) =1 + 2 cos 2\(\alpha\)

∴ L.H.S. = R.H.S. Proved

L.H.S.

=tan\(\theta\) + 2 tan 2\(\theta\) + 4 tan 4\(\theta\) + 8 cot 8\(\theta\)

=tan\(\theta\) + 2 tan 2\(\theta\) + 4 tan 4\(\theta\) + \(\frac 8{tan 8\theta}\)

=tan\(\theta\) + 2 tan 2\(\theta\) + 4 tan 4\(\theta\) + \(\frac 8{tan 2⋅4\theta}\)

=tan\(\theta\) + 2 tan 2\(\theta\) + 4 tan 4\(\theta\) + \(\frac 8{\frac {2 tan 4\theta}{1 - tan^24\theta}}\)

=tan\(\theta\) + 2 tan 2\(\theta\) + 4 tan 4\(\theta\) + \(\frac {8 (1 - tan^24\theta)}{2 tan 4\theta}\)

=tan\(\theta\) + 2 tan 2\(\theta\) + \(\frac {4 tan^24\theta + 4 (1 - tan^24\theta)}{tan 4\theta}\)

=tan\(\theta\) + 2 tan 2\(\theta\) + \(\frac {4 tan^24\theta + 4 - 4 tan^24\theta}{tan 4\theta}\)

=tan\(\theta\) + 2 tan 2\(\theta\) + \(\frac 4{tan 4\theta}\)

=tan\(\theta\) + 2 tan 2\(\theta\) + \(\frac 4{tan 2⋅2\theta}\)

=tan\(\theta\) + 2 tan 2\(\theta\) + \(\frac 4{\frac {2 tan 2\theta}{1 - tan^22\theta}}\)

=tan\(\theta\) + 2 tan 2\(\theta\) + \(\frac {2 (1 - tan^2\theta)}{tan 2\theta}\)

= tan\(\theta\) + \(\frac {2 tan^22\theta + 2 - 2 tan^22\theta}{tan 2\theta}\)

= tan\(\theta\) + \(\frac 2{tan 2\theta}\)

= tan\(\theta\) + \(\frac 2{\frac {2 tan\theta}{1 - tan^2\theta}}\)

= tan\(\theta\) + \(\frac {1 - tan^2\theta}{tan\theta}\)

= \(\frac {tan^2\theta + 1 - tan^2\theta}{tan\theta}\)

= \(\frac 1{tan\theta}\)

= cot\(\theta\)

Hence, L.H.S. = R.H.S. proved

R.H.S.

= 8 (cos2x - cos2y) (cos2x - sin2y)

= 2 (2 cos2x - 2 cos2y) (2 cos2x - 2 sin2y)

= 2 [1 + cos 2x - (2 + cos 2y)] [1 + cos 2x - (1 - cos 2y)]

= 2 [1 + cos 2x - 2 - cos 2y] [1 + cos 2x - 1 + cos 2y]

= 2 (cos 2x - cos 2y) (cos 2x - cos 2y)

= 2 (cos22x - cos22y)

= 2 cos22x - 2 cos22y

= 1 + cos 4x - (1 + cos 4y)

= 1 + cos 4x - 1 - cos 4y

= cos 4x - cos 4y

Hence, L.H.S. =R.H.S. Proved

R.H.S.

= \(\frac 2{\sqrt {2 + \sqrt {2 + 2 cos 4x}}}\)

= \(\frac 2{\sqrt {2 + \sqrt {2 + 2 cos 2 ⋅2x}}}\)

=\(\frac 2{\sqrt {2 + \sqrt {2 (1 + cos 2 ⋅2x)}}}\)

=\(\frac 2{\sqrt {2 + \sqrt {2 ⋅ 2 cos^22x}}}\)

= \(\frac 2{\sqrt {2 + 2 cos 2x}}\)

=\(\frac 2{\sqrt {2 (1 + cos 2x)}}\)

=\(\frac 2{\sqrt {2 ⋅ 2 cos^2x}}\)

= \(\frac 2{2 cosx}\)

= \(\frac 1{cosx}\)

= sec x

Hence, L.H.S. = R.H.S. Proved

L.H.S.

=4 sin3\(\alpha\)⋅ cos 3\(\alpha\) + 4 cos3\(\alpha\) ⋅sin 3\(\alpha\)

= (3 sin\(\alpha\) - sin3\(\alpha\))⋅ cos 3\(\alpha\) + (3 cos\(\alpha\) + cos 3\(\alpha\))⋅ sin 3\(\alpha\)

= 3 sin\(\alpha\) cos 3\(\alpha\) - sin 3\(\alpha\) cos 3\(\alpha\) + 3 cos\(\alpha\) sin 3\(\alpha\) + sin 3\(\alpha\) cos 3\(\alpha\)

= 3 sin\(\alpha\) cos 3\(\alpha\) + 3 cos\(\alpha\) sin 3\(\alpha\)

= 3 (sin\(\alpha\) cos 3\(\alpha\) + cos\(\alpha\) sin 3\(\alpha\))

= 3 sin(\(\alpha\) + 3\(\alpha\))

= 3 sin 4\(\alpha\)

Hence, L.H.S. = R.H.S. Proved

L.H.S.

=cos3\(\alpha\) . cos 3\(\alpha\) + sin3\(\alpha\) . sin 3\(\alpha\)

= cos3\(\alpha\) (4 cos3\(\alpha\) - 3 cos\(\alpha\)) + sin3\(\alpha\) (3 sin\(\alpha\) - 4 sin3\(\alpha\))

= 4 cos6\(\alpha\) - 3 cos4\(\alpha\) + 3 sin4\(\alpha\) - 4 sin6\(\alpha\)

= 4 (cos6\(\alpha\) - sin6\(\alpha\)) - 3 (cos4\(\alpha\) - sin4\(\alpha\))

= 4 {(cos2\(\alpha\))3 - (sin2\(\alpha\))3} - 3{(cos2\(\alpha\))2 - (sin2\(\alpha\))2}

= 4 (cos2\(\alpha\) - sin2\(\alpha\)) (cos4\(\alpha\) + cos2\(\alpha\) sin2\(\alpha\) + sin4\(\alpha\)) - 3 (cos2\(\alpha\) +sin2\(\alpha\))(cos2\(\alpha\) - sin2\(\alpha\))

= 4 cos 2\(\alpha\) {(cos2\(\alpha\))2 + (sin2\(\alpha\))2 + cos2\(\alpha\) sin2\(\alpha\)} - 3 cos 2\(\alpha\)

= cos 2\(\alpha\) [4 {(cos2\(\alpha\) + sin2\(\alpha\))2 - 2 cos2\(\alpha\) sin2\(\alpha\) + cos2\(\alpha\) sin2\(\alpha\)} - 3]

= cos 2\(\alpha\) [4 - 4 cos2\(\alpha\) sin2\(\alpha\) - 3]

= cos 2\(\alpha\) [1 - sin22\(\alpha\)]

= cos 2\(\alpha\)⋅ cos22\(\alpha\)

= cos32\(\alpha\)

∴ L.H.S. = R.H.S. Proved

R.H.S.

=2 (cos 4\(\theta\) + 1)

= 2 [cos 2. 2\(\theta\)] + 1

= 2 [2 cos2 2\(\theta\) - 1] + 1

= 4 cos2 2\(\theta\) - 2 + 1

= 4 (2 cos2\(\theta\) - 1)2 - 1

= 4 (4 cos4\(\theta\) - 4 cos2\(\theta\) + 1) - 1

= 16 cos4\(\theta\) - 16 cos2\(\theta\) + 4 - 1

= 16 cos4\(\theta\) - 16 cos2\9\theta\) + 3

= 16 cos4\(\theta\) - 12 cos2\(\theta\) - 4 cos2\(\theta\) + 3

= 4 cos2\(\theta\) (4 cos2\(\theta\) - 3) - 1 (4 cos2\(\theta\) - 3)

= (4 cos2\(\theta\) - 3) (4 cos2\(\theta\) - 1)

=(4 cos2\(\theta\) - 2 - 1) [(2 cos\(\theta\))2 - (1)2]

= [2 (2 cos2\(\theta\) -1) - 1] (2 cos\(\theta\) + 1) (2 cos\(\theta\) - 1)

= (2 cos 2\(\theta\) - 1)(2 cos\(\theta\) + 1) (2 cos\(\theta\) - 1)

=(2 cos\(\theta\) + 1) (2 cos\(\theta\) - 1)(2 cos 2\(\theta\) - 1)

Hence, L.H.S. = R.H.S. Proved

L.H.S.

=2 (cos6x + sin6x) - 3 (cos4x + sin4x) + 1

= 2 [(cos2x)3 + (sin2x)3] - 3 [(cos2x)2 + (sin2x)2] + 1

= 2 (cos2x + sin2x) [(cos2x)2 - cos2x sin2x + (sin2x)2] - 3 [(cos2x + sin2x)2 - 2 cos2x sin2x] + 1

= 2× 1 [(cos2x + sin2x)2 - 2 cos2x sin2x - cos2x sin2x] - 3 [1 - 2 sin2x cos2x] + 1

= 2 (1 - 3 cos2x sin2x) - 3 + 6 cos2x sin2x + 1

= 2 - 6 cos2x sin2x - 2 +6 cos2x sin2x

= 0

Hence, L.H.S. = R.H.S. Proved

R.H.S.

= tan (A + B)

= \(\frac {sin (A + B)}{cos (A + B)}\)× \(\frac {sin (A - B)}{sin (A - B)}\)

= \(\frac {(sinA cosB + cosA sinB) (sinA cosB - cosA sinB)}{(cosA cosB - sinA sinB) (sinA cosB - cosA sinB)}\)

= \(\frac {sin^2A cos^2B - cos^2A sin^2B}{sinA cosA cos^2B - sinB cosB cos^2A - sinB cosB sin^2A - cosA sinA sin^2B}\)

= \(\frac {sin^2A (1 - sin^2B) - (1 - sin^2A) sin^2B}{sinA cosA (cos^2B + sin^2B) - sinB cosB (cos^2A + sin^2A)}\)

= \(\frac {sin^2A - sin^2A sin^2B - sin^2B + sin^2A sin^2B}{sinA cosA . 1 - sinB cosB . 1}\)

=\(\frac {sin^2A - sin^2B}{sinA cosA - sinB cosB}\)

Hence, L.H.S. = R.H.S. Proved

L.H.S.

=cos6\(\theta\) - sin6\(\theta\)

= (cos2\(\theta\))3- (sin2\(\theta\))3

= (cos2\(\theta\) - sin2\(\theta\)) [(cos2\(\theta\))2 + cos2\(\theta\) sin2\(\theta\) + (sin2\(\theta\))2]

= cos 2\(\theta\) [(cos2\(\theta\))2 + 2 cos2\(\theta\) sin2\(\theta\) + (sin2\(\theta\))2 - cos2\(\theta\) sin2\(\theta\)]

= cos 2\(\theta\) [(cos2\(\theta\) + sin2\(\theta\))2 - \(\frac 14\) × 4 cos2\(\theta\) sin2\(\theta\)]

= cos 2\(\theta\) [(1)2 - \(\frac 14\) (2 sin\(\theta\) cos\(\theta\))2]

= cos 2\(\theta\) [1 - \(\frac 14\) (sin 2\(\theta\))2]

=cos 2\(\theta\) (1 - \(\frac 14\) sin2 2\(\theta\))

Hence, L.H.S. = R.H.S. Proved

0%
  • If sin A = (frac{3}{4}), find the value of cos 2A.                                                                                                                                                                                                                  

    -(frac{1}{8})


    - (frac{1}{-1})


    - (frac{1}{8})


    - (frac{8}{8})


  • If 5sinθ = 4 , find the value of cos 2θ.

     (frac{-7}{15})


     (frac{7}{25})


     - (frac{7}{25})


     - (frac{8}{23})


  • If cosθ = (frac{3}{4}) , find the value of sin 2θ 

    (frac{-3 sqrt{-7}}{-3})


    (frac{sqrt{5}}{2})


    (frac{3 sqrt{7}}{8})


    (frac{3 sqrt{16}}{7})


  • If cosθ = (frac{12}{13}) and sinα = (frac{4}{5}) , find the value f 2θ  and cos 2α .

    (frac{220}{069}) , (frac{-7}{25})


    (frac{100}{169}) , (frac{25}{-7})


    (frac{10}{19}) , (frac{-25}{7})


    (frac{120}{169}) , (frac{-7}{25})


  • If cosθ = (frac{3}{5}) , find the value of cos 3θ . 

    (frac{-117}{125})


    (frac{125}{115})


    (frac{117}{-125})


    (frac{127}{150})


  • If cosθ = (frac{4}{5}) , find the value of cos3θ .

     (frac{-44}{125})


     - (frac{3}{15})


     - (frac{44}{125})


     (frac{24}{144})


  • If sinα = (frac{3}{4}) , find the value f cos 2α and sinα . 

    (frac{8}{1}) , (frac{9}{16})


    (frac{9}{8}) , (frac{19}{10})


     (frac{-1}{-8}) , (frac{9}{-16})


     - (frac{1}{8}) , (frac{9}{16})


  • If tan (frac{A}{2}) = (frac{3}{4}) , find the numerical value of tan A.

    (frac{-21}{7})


    (frac{24}{6})


    (frac{21}{3})


    (frac{24}{7})


  • If cos (frac{A}{2}) = (frac{3}{5}) , find the value of cos A.

    (frac{-7}{-25})


    - (frac{7}{25})


    (frac{17}{15})


    - (frac{7}{21})


  • Find the value of sin 15(^o) if cos 30(^o) = (frac{sqrt{3}}{2}) .

    (frac {sqrt{1-3} }{3 sqrt{1}})


        (frac{sqrt{1-2}}{2 sqrt{0}})


        (frac{sqrt{3+1}}{2 sqrt{-2}})


        (frac{sqrt{3-1}}{2 sqrt{2}})


  • If cos 45(^o) = (frac{1}{sqrt{2}}) , prove that : ( 22 (frac{1}{2}) )(^o) =   (frac{sqrt{2-sqrt{2}}}{2}).

    21


    24


    12


    22


  • Express sin A interms of cot (frac{A}{2}).

    (frac {2cotfrac A2}{1 + cot^2frac A2})


    (frac {2cotfrac A4}{3 + cot^2frac A1})


    (frac {-2cotfrac A-2}{1 + cot^3frac A3})


    (frac {1cotfrac A1}{3 + cot^-3frac- A2})


  • If cos (frac{α}{3}) =  (frac{1}{2}) , find the value of cosα.

    -1


    -2


    1


    2


  • If sin (frac{θ}{3}) = (frac{4}{5}) , find the value of sinθ.

    (frac{30}{15})


    (frac{144}{44})


    (frac{44}{125})


    (frac{125}{44})


  • If sin (frac{θ}{3}) = (frac{1}{2}) , find the value of sinθ .

    2


    1


    3


    -2


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Rabin

(sec2A 1) (sec4A 1) (sec8A 1)=Tan8A.CotA

sajan

find 2sinA