2 Marks Questions

Question 512 Marks

If $\tan\theta=\frac{3}{4},$ find the value of $\frac{1-\cos\theta}{1+\cos\theta}$.

Answer

$\sec\theta=\sqrt{1+\tan^2\theta}$
$=\sqrt{1+\Big(\frac{3}{4}\Big)^2}=\sqrt{1+\frac{9}{16}}$
$\Rightarrow\ \sqrt{\frac{16-9}{16}}=\frac{5}{4}$
$\therefore\ \sec\theta=\frac{1}{\cos\theta}=\frac{1}{\frac{5}{4}}=\frac{4}{5}=\cos\theta$
$\therefore\ \text{We get }\frac{1-\frac{4}{5}}{1+\frac{4}{5}}=\frac{\frac{1}{5}}{\frac{9}{5}}=\frac{1}{9}$

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Question 522 Marks

If $2\sin^2\theta-\cos^2\theta=2,$ then find the value of $\theta$.

Answer

Given, $2\sin^2\theta-\cos^2\theta=2$
$\Rightarrow\ 2\sin^2\theta-(1-\sin^2\theta)=2 \ \big[\because\ \sin^2\theta+\cos^2\theta=1\big]$
$\Rightarrow\ \sin^2\theta+\sin^2\theta-1=2$
$\Rightarrow\ 3\sin^2\theta=3$
$\Rightarrow\ \sin^2\theta=1\ \big[\because\ \sin90^\circ=1\big]$
$\Rightarrow\ \sin\theta=1=\sin90^\circ$
$\therefore\ \theta=90^\circ$

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Question 532 Marks

Prove the following trigonometric identities.
If $\text{cosec }\theta+\cot\theta=\text{m and cosec }\theta-\cot\theta=\text{n},$ prove that mn = 1.

Answer

$\text{L.H.S}=\text{mn}$
$=(\text{cosec }\theta+\cot\theta)(\text{cosec }\theta-\cot\theta)$
$=\text{coese}^2\theta-\cot^2\theta$
$=1 \big[\because(\text{a}+\text{b})(\text{a}-\text{b})=\text{a}^2-\text{b}^2\text{ cosec}^2\theta-\cot^2\theta=1\big]$
$=1=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$

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Question 542 Marks

Prove the following trigonometric identities.
$\sin^2\text{A}\cos^2\text{B}-\cos^2\text{A}\sin^2\text{B}=\sin^2\text{A}-\sin^2\text{B}$

Answer

$\text{L.H.S}=\sin^2\text{A}\cos^2\text{B}-\cos^2\text{A}\sin^2\text{B}$
$=\sin^2\text{A}(1-\sin^2\text{B})-(1-\sin^2\text{A})(\sin^2\text{B})$
$(\because \cos^2\text{A}=1-\sin^2\text{A})$
$=\sin^2\text{A}-\sin^2\text{A}\sin^2\text{B}-\sin^2\text{B}+\sin^2\text{A}\sin^2\text{B}$
$=\sin^2\text{A}-\sin^2\text{B}=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$

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Question 552 Marks

Prove the following trigonometric identities.
$\frac{(1+\cot^2\theta)\tan\theta}{\sec^2\theta}=\cot\theta$

Answer

We have to prove $\frac{(1+\cot^2\theta)\tan\theta}{\sec^2\theta}=\cot\theta$
We know that, $\sec^2\theta-\tan^2\theta=1$
So,
$\text{L.H.S}=\frac{(1+\cot^2\theta)\tan\theta}{\sec^2\theta}=\frac{(1+\cot^2\theta)\tan\theta}{1+\tan^2\theta}$
$=\frac{\Big(1+\frac{1}{\tan^2\theta}\Big)\tan\theta}{(1+\tan^2\theta)}$
$=\frac{\Big(\frac{\tan^2\theta+1}{\tan^2\theta}\Big)\tan\theta}{(1+\tan^2\theta)}$
$=\frac{(1+\tan^2\theta)\tan\theta}{\tan^2\theta(1+\tan^2\theta)}$
$=\frac{1}{\tan\theta}$
$=\cot\theta=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$

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Question 562 Marks

Prove the following trigonometric identities.
$\text{cosec}^6\theta=\cot^6\theta+3\cot^2\theta\text{cosec}^2\theta+1$

Answer

$\text{L.H.S}=\text{cosec}^6\theta$
$=(\text{cosec}^2\theta)^3$
$=(1+\cot^2\theta)^3 \big[\because \text{cosec}^2\theta=1+\cot^2\theta\big]$
$=(1)^3+(\cot^2\theta)^3+3\cot^2\theta\times1(1+\cot^2\theta)$
$=1+\cot^6\theta+3\cot^2\theta\text{cosec}^2\theta$
$=\cot^6\theta+3\cot^2\theta\text{cosec}^2\theta+1=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$

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Question 572 Marks

Prove the following trigonometric identities.
$(\text{cosec}\theta+\sin\theta)(\text{cosec}\theta-\sin\theta)=\cot^2\theta+\cos^2\theta$

Answer

$\text{L.H.S}=(\text{cosec}\theta+\sin\theta)(\text{cosec}\theta-\sin\theta)$
$=(\text{cosec}^2\theta-\sin^2\theta)$
$=(1+\cot^2\theta)-(1-\cos^2\theta)$
$=1+\cot^2\theta-1+\cos^2\theta$
$=\cot^2\theta+\cos^2\theta=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$

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Question 582 Marks

Prove the following trigonometric identities.
$\frac{1+\cos\text{A}}{\sin^2\text{A}}=\frac{1}{1-\cos\theta}$

Answer

We know that $\sin^2\text{A}+\cos^2\text{A}=1$
$\sin^2\text{A}=1-\cos^2\text{A}$
$\Rightarrow\ \sin^2\text{A}=(1-\cos\text{A})(1+\cos\text{A})$
$\Rightarrow\ \text{L.H.S}=\frac{(1+\cos\text{A})}{(1-\cos\text{A})(1+\cos\text{A})}=\frac{1}{1-\cos\text{A}}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$
Hence proved.

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Question 592 Marks

Prove the following trigonometric identities.
$(1-\cos^2\text{A})\text{cosec}^2\text{A}=1$

Answer

We know $\sin^2\text{A}+\cos^2\text{A}=1$
$\text{L.H.S.} = (1-\cos^2\text{A})\text{cosec}^2\text{A}$
$\sin^2\text{A}=1-\cos^2\text{A}$
$=\ \sin^2\text{A}.\text{cosec}^2\text{A}$
$=\ \sin^2\text{A}.\frac{1}{\sin^2\text{A}}=1$
$=1= \text{R.H.S.}$
$$$\therefore\ \text{L.H.S.}=\text{R.H.S}$

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Question 602 Marks

Prove the following trigonometric identities.
$(\text{cosec}\text{A}-\sin\text{A})(\sec\text{A}-\cos\text{A})(\tan\text{A}+\cot\text{A})= 1$

Answer

$\text{L.H.S.}=(\text{cosec A}-\sin\text{A})(\sec\text{A}-\cos\text{A})(\tan\text{A}+\cot\text{A})$
$=\Big[\frac{1}{\sin\text{A}}-\sin\text{A}\Big]\Big[\frac{1}{\cos\text{A}}-\cos\text{A}\Big]\Big[\frac{\sin\text{A}}{\cos\text{A}}+\frac{\cos\text{A}}{\sin\text{A}}\Big]$
$\Big(\frac{1-\sin^2\text{A}}{\sin\text{A}}\Big)\Big(\frac{1-\cos^2\text{A}}{\cos\text{A}}\Big)\Big(\frac{\sin^2\text{A}+\cos^2\text{A}}{\sin\text{A}\cos\text{A}}\Big)$
$\frac{\cos^2\text{A}}{\sin\text{A}}\times\frac{\sin^2\text{A}}{\cos\text{A}}\times\frac{1}{\sin\text{A}\cos\text{A}}$
$\frac{\cos^2\text{A}\cdot\sin^2\text{A}}{\cos^2\text{A}\cdot\sin^2\text{A}}$
$=1=\text{R.H.S}.$

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Question 612 Marks

Prove the following trigonometric identities.
$\frac{\cos\theta}{\text{cosec }\theta+1}+\frac{\cos\theta}{\text{cosec }\theta-1}=2\tan\theta$

Answer

$\text{L.H.S}=\frac{\cos\theta}{\text{cosec }\theta+1}+\frac{\cos\theta}{\text{cosec}-1}$
$=\frac{\cos\theta(\text{cosec}\theta-1)+\cos\theta(\text{cosec}\theta+1)}{(\text{cosec}\theta+1)(\text{cosec }\theta-1)}$
$=\frac{\cos\theta\text{cosec }\theta-\cos\theta+\cos\theta\text{ cosec }\theta+\cos\theta}{\text{cosec}^2\theta-1}$
$=\frac{2\cos\theta\text{ cosec }\theta}{\cot^2\theta}$
$=\frac{2\cos\theta}{\sin\theta}\times\frac{\sin^2\theta}{\cos^2\theta}$
$=2\frac{\sin\theta}{\cos\theta}$
$=2\tan\theta=\text{R.H.S}$
$\therefore\ \text{L.H.S}=\text{R.H.S}$

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MATHS 2 Marks Questions