Note: Trigonometrical Ratios of complementary angles:

For $\theta < 90^o$

(i)  $\sin (90^o - \theta) = \cos \theta$          (ii) $\cos (90^o - \theta) = \sin \theta$

(iii) $\tan (90^o - \theta) = \cot \theta$         (iv) $\cot (90^o - \theta) = \tan \theta$

(v)  $\sec (90^o - \theta) = \mathrm{cosec} \theta$      (vi) $\mathrm{cosec} (90^o - \theta) = \sec \theta$

Evaluate:

Question 1:   $\frac{\cos 22^o}{\sin 68^o}$

$\frac{\cos 22^o}{\sin 68^o} = \frac{\cos (90^o-68^o)}{\sin 68^o} = \frac{\sin 68^o}{\sin 68^o}$ $= 1$

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Question 2: $\frac{\tan 47^o}{\cot 43^o}$

$\frac{\tan 47^o}{\cot 43^o} = \frac{\tan (90^o-43^o)}{\cot 43^o} = \frac{\cot 43^o}{\cot 43^o}$ $= 1$

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Question 3:   $\frac{\sec 75^o}{\mathrm{cosec} 15^o}$

$\frac{\sec 75^o}{\mathrm{cosec} 15^o} = \frac{\sec (90^o-15^o)}{\mathrm{cosec} 5^o} = \frac{\mathrm{cosec} 15^o}{\mathrm{cosec} 15^o}$ $= 1$

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Question 4:   $\frac{\cos 55^o}{\sin 35^o}+\frac{\cot 35^o}{\tan 55^o}$

$\frac{\cos 55^o}{\sin 35^o}+\frac{\cot 35^o}{\tan 55^o}$  $=$ $\frac{\cos (90-35)}{\sin 35^o}+\frac{\cot (90-35)^o}{\tan 55^o}$  $=$ $\frac{\sin 35^o}{\sin 35^o}+\frac{\tan 55^o}{\tan 55^o}$  $= 2$

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Question 5:   $\cos^2 40^o + \cos^2 50^o$

$\cos^2 40^o + \cos^2 50^o$  $= (\cos (90^o-50^o))^2 + \cos^2 50^o = \sin^2 50^o + \cos^2 50^o = 1$

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Question 6:   $\sec^2 18^o- \cot^2 72^o$

$\sec^2 18^o- \cot^2 72^o$  $= (\sec (90^o-72^o))^2 - \cot^2 72^o = \mathrm{cosec}^2 72^o - \cot^2 72^o = 1$

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Question 7:   $\sin 15^o . \cos 75^o + \cos 15^o.\sin 75^o$

$\sin 15^o . \cos 75^o + \cos 15^o.\sin 75^o$

$= \sin (90^o - 75^o) . \cos 75^o + \cos (90^o - 75^o).\sin 75^o$

$= \cos^2 75 + \sin^2 75 = 1$

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Question 8:   $\sin 42^o . \sin 48^o - \cos 42^o.\cos 48^o$

$\sin 42^o . \sin 48^o - \cos 42^o.\cos 48^o$

$= \sin 42^o . \sin (90^o - 42^o) - \cos 42^o.\cos (90^o - 42^o)$

$= \sin 42^o.\cos 42^o - \cos 42^o. \sin 42^o = 0$

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Question 9:   $\sin (90^o-A) \cos A + \cos (90^o-A) \sin A$

$\sin (90^o-A) \cos A + \cos (90^o-A) \sin A = \cos^2 A + \sin^2 A = 1$

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Question 10:   $\sin^2 35^o + \sin^2 55^o$

$\sin^2 35^o + \sin^2 55^o = \{\sin (90-55)\}^2 + \sin^2 55^o$  $= \cos^2 55^o + \sin ^2 55^o = 1$

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Question 11:   $\frac{\cot 54^o}{\tan 36^o}+\frac{\tan 20^o}{\cot 70^o}$ $- 2$

$\frac{\cot 54^o}{\tan 36^o}+\frac{\tan 20^o}{\cot 70^o}$  = $\frac{\cot (90^o - 36^o)}{\tan 36^o}+\frac{\tan (90^o - 70^o)}{\cot 70^o}$  = $\frac{\tan 36^o}{\tan 36^o}+\frac{\cot 70^o}{\cot 70^o}$ $= 2$

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Question 12:   $\frac{2 \tan 53^o}{\cot 37^o}-\frac{\cot 80^o}{\tan 10^o}$   [2006]

$\frac{2 \tan 53^o}{\cot 37^o}-\frac{\cot 80^o}{\tan 10^o}$  = $\frac{2 \tan (90^o - 37^o)}{\cot 37^o}-\frac{\cot (90^o - 10^o)}{\tan 10^o}$  = $\frac{2 \cot 37^o}{\cot 37^o}-\frac{\tan 10^o}{\tan 10^o}$ $= 0$

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Question 13:   $\cos^2 25^o + \cos^2 65^o - \tan^2 45^o$

$\cos^2 25^o + \cos^2 65^o - \tan^2 45^o$

$= \cos^2 (90^o-65^o) + \cos^2 65^o - \tan^2 45^o$

$= \sin^2 65^o + \cos^2 65^o - \tan^2 45^o = 0$

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Question 14:   $\frac{\cos^2 32^o + \cos^2 58^o}{\sin^2 59^o + \sin^2 31^o}$

$\frac{\cos^2 32^o + \cos^2 58^o}{\sin^2 59^o + \sin^2 31^o}$  = $\frac{\cos^2 (90^o - 58^o) + \cos^2 58^o}{\sin^2 (90^o - 31^o) + \sin^2 31^o}$  = $\frac{\sin^2 58^o + \cos^2 58^o}{\cos^2 31^o + \sin^2 31^o}$ $= 1$

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Question 15: $(\frac{\sin 77^o}{\cos 13^o})^2 + (\frac{\cos 77^o}{\sin 13^o})^2$ $- 2 \cos^2 45^o$

$(\frac{\sin 77^o}{\cos 13^o})^2 + (\frac{\cos 77^o}{\sin 13^o})^2$ $- 2 \cos^2 45^o$

$(\frac{\sin (90^o - 13^o)}{\cos 13^o})^2 + (\frac{\cos (90^o - 13^o)}{\sin 13^o})^2$ $- 2 \cos^2 45^o$

$(\frac{\cos 13^o}{\cos 13^o})^2 + (\frac{\sin 13^o}{\sin 13^o})^2$ $- 2 \cos^2 45^o$

$= 1 + 1 - 2 \times$ $(\frac{1}{\sqrt{2}})^2$ $= 1$

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Question 16:   $\cos^2 26^o + \cos 64^o.\sin 26^o +$ $\frac{\tan 36^o}{\cot 54^o}$    [2012]

$\cos^2 26^o + \cos 64^o.\sin 26^o +$ $\frac{\tan 36^o}{\cot 54^o}$

= $\cos^2 26^o + \cos (90^o - 26^o).\sin 26^o +$ $\frac{\tan (90^o - 54^o)}{\cot 54^o}$

= $\cos^2 26^o + \sin^2 26^o +$ $\frac{\cot 54^o}{\cot 54^o}$

$= 2$

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Question 17:   $3$ $.\frac{\sin 72^o}{\cos 18^o} - \frac{\sec 32^o}{\mathrm{cosec} 58^o}$

$3$ $.\frac{\sin 72^o}{\cos 18^o} - \frac{\sec 32^o}{\mathrm{cosec} 58^o}$

$3$ $.\frac{\sin (90^o - 18^o)}{\cos 18^o} - \frac{\sec (90^o - 32^o)}{\mathrm{cosec} 58^o}$

$3$ $.\frac{\cos 18^o}{\cos 18^o} - \frac{\mathrm{cosec} 58^o}{\mathrm{cosec} 58^o}$ $= 3 - 1 = 2$

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Question 18:   $3 \cos 80^o . \mathrm{cosec} 10^o + 2 \sin 59^o.\sec 31^o$    [2013]

$3 \cos 80^o . \mathrm{cosec} 10^o + 2 \sin 59^o.\sec 31^o$

$= 3 \cos 80^o . \mathrm{cosec} (90^o - 80^o) + 2 \sin 59^o.\sec (90^o - 59^o)$

$= 3 \cos 80^o . \sec 80^o + 2 \sin 59^o . \mathrm{cosec} 59^o$

$= 3 + 2 = 5$

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Question 19:   $\frac{\sin 80^o}{\cos 10^o}$ $+ \sin 59^o . \sec 31^o$    [2007]

$\frac{\sin 80^o}{\cos 10^o}$ $+ \sin 59^o . \sec 31^o$

$=$ $\frac{\sin (90^o - 10^o)}{\cos 10^o}$ $+ \sin 59^o . \sec (90^o - 59^o)$

$=$ $\frac{\cos 10^o}{\cos 10^o}$ $+ \sin 59^o . \mathrm{cosec} 59^o$ $= 2$

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Question 20:   $\tan (55^o-A) - \cot (35^o+A)$

$\tan (55^o-A) - \cot (35^o+A)$

$= \tan ((90^o -(35^o+A)) - \cot (35^o+A)$

$= \cot (35^o+A) - \cot (35^o+A) = 0$

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Question 21:   $\mathrm{cosec} (65^o+A) - \sec (25^o-A)$

$\mathrm{cosec} (65^o+A) - \sec (25^o-A)$

$= \mathrm{cosec} (90^o - (25^o-A)) - \sec (25^o-A)$

$= \sec (25^o-A) - \sec (25^o-A) = 0$

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Question 22:   $2$ $.\frac{\tan 57^o}{\cot 33^o} - \frac{\cot 70^o}{\tan 20^o}$ $- \sqrt{2} \cos 45^o$

$2$ $.\frac{\tan 57^o}{\cot 33^o} - \frac{\cot 70^o}{\tan 20^o}$ $- \sqrt{2} \cos 45^o$

= $2$ $.\frac{\tan (90^o - 33^o)}{\cot 33^o} - \frac{\cot (90^o - 20^o)}{\tan 20^o}$ $- \sqrt{2} \cos 45^o$

$2$ $.\frac{\cot 33^o}{\cot 33^o} - \frac{\tan 20^o}{\tan 20^o}$ $- \sqrt{2} \cos 45^o$

$= 2 - 1 - \sqrt{2} \times$ $\frac{1}{\sqrt{2}}$

$= 0$

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Question 23: $\frac{\cot^2 41^o}{\tan^2 49^o}$ $- 2$ $\frac{\sin^2 75^o}{\cos^2 15^o}$

$\frac{\cot^2 41^o}{\tan^2 49^o}$ $- 2$ $\frac{\sin^2 75^o}{\cos^2 15^o}$

$\frac{\cot^2 (90^o - 49^o)}{\tan^2 49^o}$ $- 2$ $\frac{\sin^2 (90^o - 15^o)}{\cos^2 15^o}$

$\frac{\tan^2 49^o}{\tan^2 49^o}$ $- 2$ $\frac{\cos^2 15^o}{\cos^2 15^o}$ $=-1$

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Question 24: $\frac{\cos 70^o}{\sin 20^o} + \frac{\cos 59^o}{\sin 31^o}$ $- 8 \sin^2 30^o$

$\frac{\cos 70^o}{\sin 20^o} + \frac{\cos 59^o}{\sin 31^o}$ $- 8 \sin^2 30^o$

= $\frac{\cos (90^o - 20^o)}{\sin 20^o} + \frac{\cos (90^o - 31^o)}{\sin 31^o}$ $- 8 \sin^2 30^o$

$\frac{\sin 20^o}{\sin 20^o} + \frac{\sin 31^o}{\sin 31^o}$ $- 8 \times \frac{1}{4}$

$= 0$

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Question 25: $14 \sin 30^o + 6 \cos 60^o - 5 \tan 45^o$    [2004]

$14 \sin 30^o + 6 \cos 60^o - 5 \tan 45^o$

$= 14 \sin (90^o - 60^o) + 6 \cos 60^o - 5 \tan 45^o$

$= 14 \cos 60^o + 6 \cos 60^o - 5 \tan 45^o$

$= 20 \cos 60^o - 5 \tan 45^o$

$= 20 \times \frac{1}{2} - 5 \times 1$

$= 10-5 = 5$

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Question 26: Show that: $\tan 10^o \tan 15^o \tan 75^o \tan 80^o = 1$

$\tan 10^o \tan 15^o \tan 75^o \tan 80^o$

$= \tan (90^o - 80^o) \tan (90^o - 75^o) \tan 75^o \tan 80^o$

$= \cot 80^o \cot 75^o \tan 75^o \tan 80^o$

$= 1$. Hence proved.

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Question 27: Show that: $\sin 42^o .\sec 48^o + \cos 42^o .\mathrm{cosec} 48^o = 2$

$\sin 42^o .\sec 48^o + \cos 42^o .\mathrm{cosec} 48^o$

$= \sin 42^o .\sec (90^o - 42^o) + \cos 42^o .\mathrm{cosec} (90^o - 42^o)$

$= \sin 42^o .\mathrm{cosec} 42^o + \cos 42^o .\sec 42^o$

$= 1 + 1 = 2.$ Hence proved.

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Question 28: Show that: $\frac{\sin 26^o}{\sec 64^o} + \frac{\cos 26^o}{\mathrm{cosec} 64^o}$ $= 2$

$\frac{\sin 26^o}{\sec 64^o} + \frac{\cos 26^o}{\mathrm{cosec} 64^o}$

$=$ $\frac{\sin 26^o}{\sec (90^o - 26^o)} + \frac{\cos 26^o}{\mathrm{cosec} (90^o - 26^o)}$

$=$ $\frac{\sin 26^o}{\mathrm{cosec} 26^o} + \frac{\cos 26^o}{\sec 26^o}$

$= \sin^2 26^o + \cos^2 26^o = 1 .$. Hence proved.

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Question 29: Express in terms of angles between $0^o$ and $45^o$: $\sin 59^o + \tan 63^o$

$\sin 59^o + \tan 63^o = \sin (90^o - 31^o) + \tan (90^o - 27^o) = \cos 31^o + \cot 27^o$

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Question 30: Express in terms of angles between $0^o$ and $45^o$$\mathrm{cosec} 68^o + \cot 72^o$

$\mathrm{cosec} 68^o + \cot 72^o = \mathrm{cosec} (90^o - 32^o) + \cot (90^0 - 28^o) = \sec 32^o + \tan 28^o$

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Question 31: Express in terms of angles between $0^o$ and $45^o$$\cos 74^o + \sec 67^o$

$\cos 74^o + \sec 67^o = \cos (90^o - 16^o) + \sec (90^o - 23^o) = \sin 16^o + \mathrm{cosec} 23^o$

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Question 32: $\frac{\sin A}{\sin(90^o-A)} + \frac{\cos A}{\cos(90^o-A)}$ $= \sec A \mathrm{cosec} A$

LHS $=$ $\frac{\sin A}{\sin(90^o-A)} + \frac{\cos A}{\cos(90^o-A)}$

$=$ $\frac{\sin A}{\cos A} + \frac{\cos A}{\sin A}$

$=$ $\frac{\sin^2 A + \cos^2 A}{\sin A. \cos A}$

$= \sec A \mathrm{cosec} A$ = RHS. Hence proved.

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Question 33: $\sin A \cos A -$ $\frac{\sin A \cos(90^o-A) \cos A}{\sec(90^o - A)} - \frac{\cos A \sin(90^o-A) \sin A}{\mathrm{cosec} (90^o - A)}$ $= 0$

LHS = $\sin A \cos A -$ $\frac{\sin A \cos(90^o-A) \cos A}{\sec(90^o - A)} - \frac{\cos A \sin(90^o-A) \sin A}{\mathrm{cosec} (90^o - A)}$

$\sin A \cos A -$ $\frac{\sin^2 A \cos A}{\mathrm{cosec} A} - \frac{\cos^2 A \sin A}{\sec A}$

$= \sin A \cos A -$ $\sin^3 A \cos A - \cos^3 A \sin A$

$= \sin A \cos A - \sin A \cos A(\sin^2 A + \cos^2 A)$

$= \sin A \cos A - \sin A \cos A = 0 =$ RHS. Hence proved.

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Question 34: For $\triangle ABC$, show that: $\sin (\frac{A + B}{2}) = \cos (\frac{C}{2})$

LHS $= \sin (\frac{A + B}{2}) = \sin (\frac{180 - C}{2}) = \sin (90 - \frac{C}{2}) = \cos \frac{C}{2} =$ RHS. Hence proved.

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Question 35:  For $\triangle ABC$, show that: $\tan (\frac{B + C}{2}) = \cot (\frac{A}{2})$

LHS $= \tan (\frac{B+C}{2}) = \tan (\frac{180 - A}{2}) = \tan (90 - \frac{A}{2}) = \cot \frac{A}{2} =$ RHS. Hence proved.

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Question 36: In $\triangle ABC$ is the right angles at B. Find the value of: $\frac{\sec A . \mathrm{cosec} A - \tan A . \cot C}{\sin B}$

$\frac{\sec A . \mathrm{cosec} A - \tan A . \cot C}{\sin B}$

$=$ $\frac{\sec A . \mathrm{cosec} (90^0 -A) - \tan A . \cot (90^0 -A)}{\sin 90^o}$

$=$ $\frac{\sec A . \sec A - \tan A . \tan A}{1}$

$= \sec^2 A - \tan^2 A$

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

$= 1$

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Question 37: Find $x$ if: $\sin x = \sin 60^o \cos 30^o - \cos 60^o \sin 30^o$

$\sin x = \sin 60^o \cos 30^o - \cos 60^o \sin 30^o$

$\Rightarrow \sin x = \sin 60^o \cos (90^o - 60^o) - \cos 60^o \sin (90^o - 60^o)$

$\Rightarrow \sin x = \sin^2 60^o - \cos^2 60^o$

$\Rightarrow \sin x =$ $(\frac{\sqrt{3}}{2})^2 - (\frac{1}{2})^2$

$\Rightarrow \sin x = \frac{1}{2} = \sin 30^o$

$\Rightarrow x = 30^o$

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Question 38: Find $x$ if: $\sin x = \sin 60^o \cos 30^o + \cos 60^o \sin 30^o$

$\sin x = \sin 60^o \cos 30^o + \cos 60^o \sin 30^o$

$\Rightarrow \sin x = \sin 60^o \cos (90^ - 60^o) + \cos 60^o \sin (90^ - 60^o)$

$\Rightarrow \sin x = \sin^2 60^o + \cos^2 60^o$

$\Rightarrow \sin x = 1$

$\Rightarrow x = 90^o$

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Question 39: Find $x$ if: $\cos x = \cos 60^o \cos 30^o - \sin 60^o \sin 30^o$

$\cos x = \cos 60^o \cos 30^o - \sin 60^o \sin 30^o$

$\Rightarrow \cos x = \cos 60^o \cos (90^o - 60^o) - \sin 60^o \sin (90^o - 60^o)$

$\Rightarrow \cos x = \cos 60^o \sin 60^o - \sin 60^o \cos 30^o$

$\Rightarrow \cos x = = 0$

$\Rightarrow x = 90^o$

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Question 40: Find the value of $x$ if: $\tan x =$ $\frac{\tan 60^o - \tan 30^o}{1+ \tan 60^o \tan 30^o}$

$\tan x =$ $\frac{\tan 60^o - \tan 30^o}{1+ \tan 60^o \tan 30^o}$

$\Rightarrow \tan x =$ $\frac{\sin 60^o. \cos 30 - \sin 30^o. \cos 60}{\cos 60.\cos 30+ \sin 60^o. \sin 30^o}$

$\Rightarrow \tan x =$ $\frac{\sin 60^o. \cos (90^o - 60^o) - \sin (90^o - 60^o). \cos 60}{\cos 60.\cos 30+ \sin 60^o. \sin 30^o}$

$\Rightarrow \tan x =$ $\frac{\sin^2 60^o - \cos^2 60}{\cos 60.\cos 30+ \sin 60^o. \sin 30^o}$

$\Rightarrow \tan x =$ $\frac{(\frac{\sqrt{3}}{2})^2 - (\frac{1}{2})^2}{2 \times \frac{1}{2} \times \frac{\sqrt{3}}{2}}$

$\Rightarrow \tan x =$ $\frac{2}{\sqrt{3}}$

Therefore $x = 30^o$

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Question 41: Find the value of $x$ if: $\sin 2x = 2 \sin 45^o \cos 45^o$

$\sin 2x = 2 \sin 45^o \cos 45^o$

$\Rightarrow \sin 2x = 2 \times \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}}$

$\Rightarrow \sin 2x = 1$

$\Rightarrow 2x = 90^o \Rightarrow x = 45^o$

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Question 42: Find the value of $x$ if: $\sin 3x = 2 \sin 30^o \cos 30^o$

$\sin 3x = 2 \sin 30^o \cos 30^o$

$\Rightarrow \sin 3x = 2 \times \frac{1}{2} \times \frac{\sqrt{3}}{2}$

$\Rightarrow \sin 3x = \frac{\sqrt{3}}{2}$

$\Rightarrow 3x = 60^o \Rightarrow x = 20^o$

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Question 43: Find $x$ if: $\cos (2x-6^o) = \cos^2 30^o - \cos^2 60^o$

$\cos (2x-6^o) = \cos^2 30^o - \cos^2 60^o$

$\Rightarrow \cos (2x-6^o) = \sin^2 60^o - \cos^2 60^o$

$\Rightarrow \cos (2x-6^o) = (\frac{\frac{\sqrt{3}}{2}}{2})^2 - (\frac{1}{2})^2$

$\Rightarrow \cos (2x-6^o) = \frac{1}{2}$

$\Rightarrow 2x - 6 = 60 \Rightarrow x = 33^o$

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Question 44: Find the value of $A$ where  $0^o \leq A \leq 90^o$$\sin (90^o - 3A) . \mathrm{cosec} 42^o = 1$

$\sin (90^o - 3A) . \mathrm{cosec} 42^o = 1$

$\Rightarrow \sin (90^o - 3A) = \sin 42^o$

$\Rightarrow 90^o - 3A = 42^o \Rightarrow A = 16^o$

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Question 45: Find the value of $A$ where  $0^o \leq A \leq 90^o$$\cos (90^o - A) . \sec 77^o = 1$

$\cos (90^o - A) . \sec 77^o = 1$

$\Rightarrow \cos (90^o - A) = \cos 77^o$

$\Rightarrow (90^o - A) = 77^o \Rightarrow A = 13^o$

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Question 46: Prove that: $\frac{\cos (90^o-A).\cos A}{\cot A}$ $= 1 - \cos^2 A$

LHS $=$ $\frac{\cos (90^o-A).\cos A}{\cot A}$

$=$ $\frac{\sin A.\cos A . \sin A}{\cos A}$

$= \sin^2 A = 1- \cos^2 A =$ RHS. Hence proved.

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Question 47: Prove that: $\frac{\sin A . \sin (90^o-A)}{\tan (90^o-A)}$ $= 1 - \sin^2 A$

LHS $=$ $\frac{\sin A. \sin (90^o-A)}{\tan A}$

$=$ $\frac{\sin A.\cos A . \cos A}{\sin A}$

$= \cos^2 A = 1- \sin^2 A =$ RHS. Hence proved.

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Question 48: Evaluate: $\frac{\sin 35^o.\cos 55^o + \cos 35^o. \sin 55^o}{\mathrm{cosec}^2 10^o - \tan^2 80^o}$    [2010]

$\frac{\sin 35^o.\cos 55^o + \cos 35^o. \sin 55^o}{\mathrm{cosec}^2 10^o - \tan^2 80^o}$

$=$ $\frac{\sin 35^o.\cos (90^o - 35^o) + \cos 35^o. \sin (90^o - 35^o)}{\mathrm{cosec}^2 10^o - \tan^2 (90^o - 10^o) }$

$=$ $\frac{\sin^2 35^o + \cos^2 35^o}{1 - \cos^2 10}$ $\times \sin^2 10$

$=$ $\frac{\sin^2 10}{1- \cos^2 A}$ $= 1$

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Question 49: Without u\sing trigonometric tables, evaluate

$\sin^2 34^o + \sin^2 56^o + 2 \tan 18^o. \tan 72^o - \cot^2 30^o$    [2014]

$\sin^2 34^o + \sin^2 56^o + 2 \tan 18^o. \tan 72^o - \cot^2 30^o$
$= \sin^2 34^o + \cos^2 34^o + 2 \tan 18^o. \cot 18^o - \cot^2 30^o$
$= 3 - (\sqrt{3})^2 = 0$