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Trigonometric Ratios Of Compound Angles · Rajasthan Board (RBSE) STD 11 Science (Rajasthan - English Medium). 16 questions.

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Question 11 Mark
If $\pi<\text{x}<\frac{3\pi}{2},$ then write the value of $\sqrt{\frac{1-\cos2\text{x}}{1+\cos2\text{x}}}$
Answer
$\pi<\theta<\frac{3\pi}{2}\Rightarrow\theta$ lies in the 3rd quadarant
$\sqrt{\frac{1-\cos2\theta}{1+\cos2\theta}}$
$=\sqrt{\frac{2\sin^2\theta}{2\cos^2\theta}}$
$=\frac{-\sin\theta}{-\cos\theta}$ [$\because\theta$ lies in the 3rd quadarant]
$=\tan\theta$
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Question 21 Mark
Prove that:
$\frac{\sin2\text{x}}{1-\cos2\text{x}}=\cot\text{x}$
Answer
$\text{LHS}=\frac{\sin2\text{x}}{1-\cos\text{x}}=\frac{2\sin\text{x}\cos\text{x}}{2\sin^2\text{x}}$
$=\frac{\cos\text{x}}{\sin\text{x}}$
$=\cot\text{x}=\text{RHS}$
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Question 31 Mark
Write the angled triangle ABC, write value of $\sin^2\text{A}\sin^2\text{B}+\sin^2\text{C}.$
Answer
$\cos^276^\circ+\cos^216^\circ-\cos76^\circ\cos16^\circ$
$=\frac{1}{2}[1+\cos2(76)^\circ+1\cos2(16)^\circ-\cos(76-16)^\circ]$
$=\frac{1}{2}\big[2-\frac{1}{2}+\cos152^\circ+\cos32^\circ-\cos92^\circ\big]$
$=\frac{1}{2}\Big[\frac{3}{2}+\cos152^\circ+2\sin\frac{92+32}{2}\sin\frac{92-32}{2}\Big]$
$=\frac{1}{2}\Big[\frac{3}{2}+\cos152^\circ+2\sin\frac{124^\circ}{2}\sin\frac{60^\circ}{2}\Big]$
$=\frac{1}{2}\Big[\frac{3}{2}+\cos152^\circ+2\sin62^\circ\sin30^\circ\Big]$
$=\frac{1}{2}\Big[\frac{3}{2}+\cos152^\circ+2\sin62^\circ\Big]$
$=\frac{1}{2}\Big[\frac{3}{2}+\cos152^\circ+2\sin(90-28)^\circ\Big]$
$=\frac{1}{2}\Big[\frac{3}{2}+\cos152^\circ+\cos28^\circ\Big]$
$=\frac{1}{2}\Big[\frac{3}{2}+2\cos\frac{152-28}{2}\cos\frac{152-28}{2}\Big]$
$=\frac{1}{2}\Big[\frac{3}{2}+2\cos90^\circ\cos\frac{124^\circ}{2}\Big]$
$=\frac{1}{2}\Big[\frac{3}{2}+2(0)\cos62^\circ\Big]$
$=\frac{3}{4}$
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Question 41 Mark
If $\frac{\pi}{2}<\text{x}<\pi,$ then wrire the value of $\sqrt{\frac{1-\cos^2\text{x}}{1+cos^2\text{x}}}.$
Answer
since $\frac{\pi}{2}<\theta<\pi\Rightarrow\theta$ (lies in the 2nd quadarand)
Now,
$\sqrt{\frac{1-\cos2\theta}{1+\cos2\theta}}$
$=\sqrt{\frac{2\sin^2\theta}{1\cos^2\theta}}$
$=\frac{\sin\theta}{1\cos\theta}$ ($\because\theta$ lies in the 2nd quadarant)
$=-\tan\theta$
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Question 51 Mark
If $\tan\text{A}=\frac{1-\cos\text{B}}{\sin\text{B}},$ then find value of $\tan2\text{A}.$
Answer
given $\tan\text{A}=\frac{1\cos\text{B}}{\sin\text{B}}=\frac{2\sin^2\frac{\text{B}}{2}}{2\sin\frac{\text{B}}{2}\cos\frac{\text{B}}{2}}=\tan\frac{\text{B}}{2}$
$\tan2\text{A}=\frac{2\tan\text{A}}{1-\tan^2\text{A}}$
Subtitute the value of tan A, we get
$\tan2\text{A}=\frac{2\tan\frac{\text{B}}{2}}{1-\tan^2\frac{\text{B}}{2}}=\tan\text{B}$
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Question 61 Mark
In a right angled tringle ABC, write the value of $\sin^2\text{A}+\sin^2\text{B}+\sin^2\text{C}.$
Answer
Suppose in $\text{ABC}\angle\text{B}=90^\circ$
$\Rightarrow\text{A+C}=\frac{\pi}{2}$
$\Rightarrow\text{A}=\frac{\pi}{2}-\text{C}$
$\Rightarrow\sin\text{A}=\sin\Big(\frac{\pi}{2}-\text{C}\Big)$
Now,
$\sin^2\text{A}+\sin^2\text{B}+\sin^2\text{C}$
$=\sin^2\text{A}+1+\cos^2\text{A}$ $\big[\because\sin\frac{\pi}{2}=1\big]$
$=1+1=2$
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Question 71 Mark
If $\sin\text{x}+\cos\text{x}=\text{a},$ find the value of $\sin^6\text{x}+\cos^6\text{x}.$
Answer
Given: $\sin\text{x}+\cos\text{x}=\text{a}$
squaring on both sides, we get
$\sin^2\text{x}+\cos^2\text{x}+2\sin\text{x}\cos\text{x}=\text{a}^2$
$\Rightarrow1+2\sin\text{x}\cos\text{x}=\text{a}^2$
$\Rightarrow\sin\text{x}\cos\text{x}=\frac{\text{a}^2-1}{2}\ .....(1)$
Now,
$\sin^6\text{x}+\cos^6\text{x}$
$=\big(\sin^2\text{x}+\cos^2\text{x}\big)^3-3\sin^2\text{x}\cos^2\text{x}(\sin^2\text{x}+\cos^2\text{x})$
$=1-3\big(\frac{\text{a}^2-1}{2}\big)^2$ [Using (1)]
$=\frac{4-3(\text{a}^2-2)^1}{4}$
Hence, the required value is $\frac{1}{4}\big[4-3(\text{a}^2-1)^2\big]$
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Question 81 Mark
If $\frac{\pi}{2}<\text{x}<\pi,$ then write the valve of $\sqrt{2+\sqrt{2+2\cos2\text{x}}}$ in the simplest from.
Answer
Since $\frac{\pi}{2}<\theta<\pi\Rightarrow\theta$ lies tn the 2nd quadarant
Now,
$\sqrt{2\pm\sqrt{2+2\cos2\theta}}$
$=\sqrt{2\pm\sqrt{2(1+\cos2\theta)}}$
$=\sqrt{2\pm\sqrt{2.2\cos^2\theta}}$
$=\sqrt{2-(2\cos\theta)}$ ($\because\theta$ lies in the 2nd quadarant)
$=\sqrt{2(1-\cos\theta)}$
$=2.2\sin^2\frac{\theta}{2}$
$=2\sin\frac{\theta}{2}$
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Question 91 Mark
If $\frac{\pi}{2}<​\text{x}<\frac{3\pi}{2},$ then write the value of $\sqrt{\frac{1+\cos2​\text{x}}{2}}$
Answer
Since $\frac{\pi}{2}<\theta<\frac{3\pi}{2}\Rightarrow2^{\text{nd}}\ \&\ 3^{\text{rd}}$ quadarant
Now,
$\sqrt{\frac{1+\cos2\theta}{2}}=\sqrt{\frac{2\cos^2\theta}{2}}$
$-\cos\theta$ (-ve sigh due to 2nd 3rd quad)
$=-\cos\theta$
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Question 101 Mark
Write the value of $\cos\frac{\pi}{7}\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}.$
Answer
$\cos\frac{\pi}{7}.\cos\frac{2\pi}{7}.\cos\frac{4\pi}{7}$
$=\frac{2.\sin\frac{pi}{7}.\cos\frac{\pi}{7}.\cos\frac{2\pi}{7}.\cos\frac{4\pi}{7}}{2\sin\frac{\pi}{7}}$ $\big[$Divide and multiply by $2\sin\frac{\pi}{7}\big]$
$=\frac{2.\sin\frac{2\pi}{7}.\cos\frac{\pi}{7}.\cos\frac{2\pi}{7}.\cos\frac{4\pi}{7}}{2.2\sin\frac{\pi}{7}}$
$=\frac{2\sin\frac{4\pi}{7}.\cos\frac{4\pi}{7}}{2.4\sin\frac{\pi}{7}}$
$=\frac{\sin\frac{8\pi}{7}}{8\sin\frac{\pi}{7}}$
$=\frac{\sin\big(\pi+\frac{\pi}{7}\big)}{8\sin\frac{\pi}{7}}$
$=\frac{-1}{8}$
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Question 111 Mark
If $\cos4\text{x}=1+\text{K}\sin^2\text{x}\cos^2\text{},$ then write the value of k.
Answer
We have,
$\cos4\text{x}=1+\text{k}\sin^2\text{x}\cos^2\text{x}\ .....(\text{i})$
$\Rightarrow\cos2.2\text{x}-\cos^22\text{x}-\sin^2\text{x}$
$=1-2\sin^22\text{x}$
$=1-2(2\sin\text{x}\cos\text{x})^2$
$=1-8\sin^2\text{x}\cos^2\text{x}\ .....(\text{ii})$
compaiing (i) & (ii), we get
$\text{k}=-8$
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Question 121 Mark
If $\sin\text{x}+\cos\text{x}=\text{a},$ find the value of $|\sin\text{x}-\cos\text{x}|.$
Answer
$\sin+\cos\text{x}=\text{a}$
squaring on both sides gives
$\sin^2\text{x}+\cos^2\text{x}+2\sin\text{x}\cos\text{x}=\text{a}^2$
$1+\sin2\text{x}=\text{a}^2$
$\sin2\text{x}=\text{a}^2-1$
If $\sin2\text{x}=\text{a}^2-1.$ then
$(\sin2\text{x}=\text{a}^2)=\sin^2\text{x}+\cos^2\text{x}-2\sin\text{x}\cos\text{x}=1(\text{a}^2-1)=2-\text{a}^2$
Take square root on both side, we get
$|\sin\text{x}-\cos\text{x}|=\sqrt{2-\text{a}^2}$
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Question 131 Mark
If $\frac{\pi}{4}<\text{x}<\frac{\pi}{2},$ then write the value of $\sqrt{1-\sin2\text{x}}$
Answer
Since, $\frac{\pi}{4}<\theta<\frac{\pi}{2}\Rightarrow\theta$ lies in the first quadrant 1st quadarant
Now,
$\sqrt{1-\sin2\theta}=\sqrt{\sin^2\theta+\cos^2\theta-2\sin\theta}$
$=\sqrt{(\sin\theta-\cos\theta)^2}$
$=\sin\theta-\cos\theta$
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Question 141 Mark
If $\tan\frac{\text{x}}{2}=\frac{\text{m}}{\text{n}},$ then write the value of $\text{m}\ \sin\text{x}+\text{n}\cos\text{x}.$
Answer
We have,
$\tan\frac{\text{x}}{2}=\frac{\text{m}}{\text{n}}$
$\Rightarrow\frac{\sin\frac{\text{x}}{2}}{\cos\frac{\text{x}}{2}}=\frac{\text{m}}{\text{n}}$
$\Rightarrow\sin\frac{\text{x}}{2}=\text{mk},\&\cos\frac{\text{x}}{2}=\text{nk}(\text{say})$
Now,
$\text{m}\sin\text{x}+\text{n}\cos\text{x}$
$=\text{m}2\sin\frac{\text{x}}{2},\cos\frac{\text{x}}{2}+\text{n}\Big(\cos^2\frac{\text{x}}{2}-\sin^2\frac{\text{x}}{2}\Big)$
$2\text{m}.\text{mk}.\text{nk}+\text{n}(\text{n}^2\text{k}^2-\text{m}^2\text{k}^2)$
$=2\text{m}^2\text{k}^2\text{n}+\text{nk}^2(\text{n}^2-\text{m}^2)$
$=\text{nk}^2(2\text{m}^2+\text{n}^2-\text{m}^2)$
$=\text{nk}^2(\text{m}^2+\text{n}^2)$
$=​​\text{n}(​\text{m}^2​\text{k}^2+​\text{n}^2​\text{k}^2)$
$​\text{n}=\Big(\sin^2\frac{​\text{x}}{2}+\cos^2\frac{​\text{x}}{2}\Big)$
$=​\text{n}$
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Question 151 Mark
Prove that:
$\sqrt{\frac{1-\cos2\text{x}}{1+\cos2\text{x}}}=\tan\text{x}$
Answer
We have,
$\text{LHS}=\sqrt{\frac{1-\cos2\text{x}}{1+\cos2\text{x}}}$
$\sqrt{\frac{1-\cos2\text{x}}{1+\cos2\text{x}}}=\sqrt{\frac{2\sin^2\text{x}}{2\cos^2\text{x}}}$
$=\frac{\sin\text{x}}{\cos\text{x}}$
$\tan\text{x}=\text{RHS.}$
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Question 161 Mark
Prove that:
$\frac{\sin2\text{x}}{1+\cos2\text{x}}=\tan\text{x}$
Answer
$\text{LHS}=\frac{\sin2\text{x}}{1+\cos2\text{x}}=\frac{2\sin\text{x},\cos\text{x}}{2\cos^2\text{x}}$
$=\frac{\sin\text{x}}{\cos\text{x}}$
$=\tan\text{x}=\text{RHS}$
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