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}

Answer:

\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}

Answer:

\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}

Answer:

\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}

Answer:

\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

Answer:

\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

Answer:

\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

Answer:

\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

Answer:

\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

Answer:

\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

Answer:

\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

Answer:

\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]

Answer:

\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

Answer:

\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}

Answer:

\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

Answer:

(\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]

Answer:

\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}

Answer:

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]

Answer:

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]

Answer:

\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)

Answer:

\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)

Answer:

\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

Answer:

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}

Answer:

\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

Answer:

\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]

Answer:

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

Answer:

\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

Answer:

\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

Answer:

\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

Answer:

\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

Answer:

\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

Answer:

\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

Answer:

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

Answer:

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})

Answer:

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})

Answer:

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}

Answer:

\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

Answer:

\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

Answer:

\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

Answer:

\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}

Answer:

\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

Answer:

\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

Answer:

\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

Answer:

\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

Answer:

\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

Answer:

\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

Answer:

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

Answer:

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]

Answer:

\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]

Answer:

\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