# Class 10: Trigonometry – Sample Problems Exercise 23(c-2)

Question 1: 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 2: Show that: $sin \ 42^o .sec \ 48^o + cos \ 42^o .cosec \ 48^o = 2$

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

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

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

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

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Question 3: Show that: $\frac{sin \ 26^o}{sec \ 64^o} + \frac{cos \ 26^o}{cosec \ 64^o}$$= 2$

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

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

$=$$\frac{sin \ 26^o}{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 4: 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 5: Express in terms of angles between $0^o$ and $45^o$$cosec \ 68^o + cot \ 72^o$

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

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Question 6: Express in terms of angles between $0^o$ and $45^o$$cos \ 74^o + sec \ 67^o$

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

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Question 7: $\frac{sin \ A}{sin\ (90^o-A)} + \frac{cos \ A}{cos\ (90^o-A)}$$= sec \ A 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}$

$= sec \ A cosec \ A$ = RHS. Hence proved.

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Question 8:

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

$sin \ A \ cos \ A -$$\frac{sin^2 \ A \ cos \ A}{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 9: 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 10:  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 11: In $\triangle ABC$ is the right angles at B. Find the value of:

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

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

$=$$\frac{sec \ A . 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 12: 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 13: 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 14: 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 15: 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 16: 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 17: 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 18: Find $x$ if: $latex cos \ (2x-6^o) = cos^2 \ 30^o – cos^2 \ 60^o &s=0$

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

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

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

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

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

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Question 19: Find the value of $A \ where \ 0^o \leq A \leq 90^o$$sin \ (90^o - 3A) . cosec \ 42^o = 1$

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

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

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

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Question 20: 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 21: 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 22: 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 23: Evaluate: $\frac{sin \ 35^o.cos \ 55^o + cos \ 35^o. sin \ 55^o}{cosec^2 \ 10^o - tan^2 \ 80^o}$    [2010]

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

$=$$\frac{sin \ 35^o.cos \ (90^o - 35^o) + cos \ 35^o. sin \ (90^o - 35^o)}{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 24: Without using trigonometric tables, evaluate

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

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

$= sin^2 \ 34^o + cos^2 \ 34^o + 2 tan \ 18^o. cot \ 18^o - cot^2 \ 30^o$

$= 3 - (\sqrt{3})^2 = 0$

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Class 10: Trigonometry – Sample Problems Exercise 23(c-2)

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