3 Marks Question

Question 13 Marks

If $\sec2\text{A}=\text{cosec}(\text{A}-42^\circ),$ where 2A is an acute angle, then find the value of A.

Answer

We have,
$\sec2\text{A}=\text{cosec}(\text{A}-42^\circ),$
$\Rightarrow\text{cosec}(90^\circ-2\text{A})=\text{cosec}(\text{A}-42^\circ)$
Comparing both sides, we get
$90^\circ-2\text{A}=\text{A}-42^\circ$
$\Rightarrow2\text{A}+\text{A}=90^\circ+42^\circ$
$\Rightarrow3\text{A}=132^\circ$
$\Rightarrow\text{A}=\frac{132^\circ}{3}$
$\therefore\text{A}=44^\circ$
Hence, the value of A is 44°.

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Question 23 Marks

Prove that:
$\tan5^\circ\tan25^\circ\tan30^\circ\tan65^\circ\tan85^\circ=\frac{1}{\sqrt{3}}$

Answer

$\text{L.H.S.}=\tan5^\circ\tan25^\circ\tan30^\circ\tan65^\circ\tan85^\circ$
$=\tan(90^\circ-85^\circ)\tan(90^\circ-65^\circ)\times\frac{1}{\sqrt{3}}\times\frac{1}{\cot65^\circ}\frac{1}{\cot85^\circ}$
$=\cot85^\circ\cot65^\circ\frac{1}{\sqrt{3}}\frac{1}{\cot65^\circ}\frac{1}{\cot85^\circ}$
$=\frac{1}{\sqrt{3}}$
$=\text{R.H.S.}$

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Question 33 Marks

If $\tan2\text{A}=\cot(\text{A}-12^\circ),$ where 2A is an acute angle, then find the value of A.

Answer

$\tan2\text{A}=\cot(\text{A}-12^\circ)$
$\Rightarrow\cot(90^\circ-2\text{A})=\cot(\text{A}-12^\circ)$ $\big[\because\ \tan\theta=\cot(90^\circ-\theta)\big]$
$\Rightarrow(90^\circ-2\text{A})=(\text{A}-12^\circ)$
$\Rightarrow102^\circ=3\text{A}$
$\Rightarrow\text{A}=\frac{102^\circ}{3}=34^\circ$

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Question 43 Marks

Prove that:
$\cot\theta\tan(90^\circ-\theta)-\sec(90^\circ-\theta)\text{cosec }\theta\\+\sqrt{3}\tan12^\circ\tan60^\circ\tan78^\circ=2$

Answer

$\text{L.H.S.}=\cot\theta\tan(90^\circ-\theta)-\sec(90^\circ-\theta)\text{cosec }\theta\\ \ \ +\sqrt{3}\tan12^\circ\tan60^\circ\tan78^\circ$
$=\cot\theta\cot\theta-\text{cosec }\theta\text{ cosec }\theta\\ \ +\sqrt{3}\tan12^\circ\times\sqrt{3}\times\cot(90^\circ-78^\circ)$
$=\cot^2\theta-\text{cosec}^2\theta+3\tan12^\circ\cot12^\circ$
$=-1+3$
$=2$
$=\text{R.H.S.}$
Hence Proved.

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Question 53 Marks

Prove that:
$\cos1^\circ\cos2^\circ\cos3^\circ\dots\cos180^\circ=0$

Answer

$\text{L.H.S.}=\cos1^\circ\cos2^\circ\cos3^\circ\dots\cos180^\circ$
$=\cos1^\circ\times\cos2^\circ\times\cos3^\circ\times\dots\cos90^\circ\times\dots\cos180^\circ$
$=\cos1^\circ\times\cos2^\circ\times\cos3^\circ\times\dots\times0\times\dots\cos180^\circ$
$=0$
$=\text{R.H.S.}$

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Question 63 Marks

If $\cos2\theta=\sin4\theta,$ where $2\theta$ and $4\theta$ are acute angles, find the value of $\theta.$

Answer

We have,
$\cos2\theta=\sin4\theta,$
$\Rightarrow\sin(90^\circ-2\theta)=\sin4\theta$
Comparing both sides, we get
$90^\circ-2\theta=4\theta$
$90^\circ-2\theta=4\theta$
$\Rightarrow2\theta+4\theta=90^\circ$
$\Rightarrow6\theta=90^\circ$
$\Rightarrow\theta=\frac{90^\circ}{6}$
$\therefore\theta=15^\circ$
Hence, the value of $\theta$ is $15^\circ.$

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Question 73 Marks

Prove that:
$\frac{\cos(90^\circ-\theta)}{1+\sin(90^\circ-\theta)}+\frac{1+\sin(90^\circ-\theta)}{\cos(90^\circ-\theta)}=2\text{cosec }\theta$

Answer

$\text{L.H.S.}=\frac{\cos(90^\circ-\theta)}{1+\sin(90^\circ-\theta)}+\frac{1+\sin(90^\circ-\theta)}{\cos(90^\circ-\theta)}$
$=\frac{\sin\theta}{1+\cos\theta}+\frac{1+\cos\theta}{\sin\theta}$
$=\frac{\sin^2\theta+(1+\cos\theta)^2}{(1+\cos\theta)\sin\theta}$
$=\frac{\sin^2+(1+\cos\theta)^2}{(1+\cos\theta)\sin\theta}$
$=\frac{\sin^2\theta+1+\cos^2\theta+2\cos\theta}{(1+\cos\theta)\sin\theta}$
$=\frac{1+1+2\cos\theta}{(1+\cos\theta)\sin\theta}$
$=\frac{2+2\cos\theta}{(1+\cos\theta)\sin\theta}$
$=\frac{2(1+\cos\theta)}{(1+\cos\theta)\sin\theta}$
$=2\frac{1}{\sin\theta}$
$=2\text{cosec }\theta$
$=\text{R.H.S.}$
Hence Proved.

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Question 83 Marks

Prove that:
$\frac{\cos(90^\circ-\theta)\sec(90^\circ-\theta)\tan\theta}{\text{cosec}(90^\circ-\theta)\sin(90^\circ-\theta)\cot(90^\circ-\theta)}+\frac{\cos\theta\sin(90^\circ-\theta)}{\cot\theta}=2$

Answer

$\text{L.H.S.}=\frac{\cos(90^\circ-\theta)\sec(90^\circ-\theta)\tan\theta}{\text{cosec}(90^\circ-\theta)\sin(90^\circ-\theta)\cot(90^\circ-\theta)}\\ \ +\frac{\cos\theta\sin(90^\circ-\theta)}{\cot\theta}$
$=\frac{\sin\theta\text{cosec }\theta\tan\theta}{\sec\theta\cos\theta\tan\theta}+\frac{\cot\theta}{\cot\theta}$
$=1+1$
$=2$
$=\text{R.H.S.}$
Hence Proved.

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Question 93 Marks

Prove that:
$\sin(70^\circ+\theta)-\cos(20^\circ-\theta)=0$

Answer

$\text{L.H.S.}=\sin(70^\circ+\theta)-\cos(20^\circ-\theta)$
$=\sin\big\{90^\circ-(20^\circ-\theta)\big\}-\cos(20^\circ-\theta)$
$=\cos(20^\circ-\theta)-\cos(20^\circ-\theta)$
$=0$
$=\text{R.H.S.}$

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Question 103 Marks

If A, B and C are the angle of a $\triangle\text{ABC},$ prove that $\tan\Big(\frac{\text{C+A}}{2}\Big)=\cot\frac{\text{B}}{2}.$

Answer

In $\triangle\text{ABC},$
$\text{A}+\text{B}+\text{C}=180^\circ$
$\Rightarrow\text{A}+\text{C}=180^\circ-\text{B}\dots(\text{i})$
Now,
$\text{L.H.S.}=\tan\Big(\frac{\text{C}+\text{A}}{2}\Big)$
$=\tan\Big(\frac{180^\circ-\text{B}}{2}\Big)$ [Using (i)]
$=\tan\Big(90^\circ-\frac{\text{B}}{2}\Big)$
$=\cot\frac{\text{B}}{2}$
$=\text{R.H.S.}$

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Question 113 Marks

If $\sec4\text{A}=\text{cosec}(\text{A}-15^\circ),$ where 4A is an acute angle, then find the value of A.

Answer

$\sec4\text{A}=\text{cosec}(\text{A}-15^\circ)$
$\Rightarrow\text{cosec}(90^\circ-4\text{A})=\text{cosec}(\text{A}-15^\circ)$ $\big[\because\ \sec\theta=\text{cosec}(90^\circ-\theta)\big]$
$\Rightarrow90^\circ-4\text{A}=\text{A}-15^\circ$
$\Rightarrow105^\circ=5\text{A}$
$\Rightarrow\text{A}=\frac{105^\circ}{5}=21^\circ$

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Question 123 Marks

If $\sin3\text{A}=\cos(\text{A}-26^\circ),$ where 3A is an acute angle, then find the value of A.

Answer

$\sin3\text{A}=\cos(\text{A}-26^\circ),$
$\Rightarrow\cos(90^\circ-3\text{A})=\cos(\text{A}-26^\circ)$ $\big[\because\ \sin\theta=\cos(90^\circ-\theta)\big]$
$\Rightarrow90^\circ-3\text{A}=\text{A}-26^\circ$
$\Rightarrow116^\circ=4\text{A}$
$\Rightarrow\text{A}=\frac{116^\circ}{4}=29^\circ$

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Question 133 Marks

Prove that:
$\Big(\frac{\sin49^\circ}{\cos41^\circ}\Big)^2+\Big(\frac{\cos41^\circ}{\sin49^\circ}\Big)^2=2$

Answer

$\text{L.H.S.}=\Big(\frac{\sin49^\circ}{\cos41^\circ}\Big)^2+\Big(\frac{\cos41^\circ}{\sin49^\circ}\Big)^2$
$=\Big(\frac{\cos(90^\circ-49^\circ)}{\cos41^\circ}\Big)^2+\Big(\frac{\cos41^\circ}{\cos(90^\circ-49^\circ)}\Big)^2$
$=\Big(\frac{\cos41^\circ}{\cos41^\circ}\Big)^2+\Big(\frac{\cos41^\circ}{\cos41^\circ}\Big)^2$
$=1^2+1^2$
$=1+1$
$=2$
$=\text{R.H.S.}$
Disclaimer: The RHS of (v) given in textbook is incorrect. There should be 2 instead 1. The same has been corrected in the solution here.

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