(Each question 1 marks) - MATHS STD 11 Science Questions - Vidyadip

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16 questions · self-marked practice — reveal the answer and mark yourself.

Question 11 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 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

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 41 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 51 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 $2^{nd}$ 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 $2^{nd}$ quadarant) $=-\tan\theta$

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Question 61 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 71 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 81 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 91 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 101 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 $2^{nd}$ 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 $2^{nd}$ quadarant) $=\sqrt{2(1-\cos\theta)}$ $=2.2\sin^2\frac{\theta}{2}$ $=2\sin\frac{\theta}{2}$

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Question 111 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 $2^{nd}$ $3^{rd}$ quad) $=-\cos\theta$

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Question 121 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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Question 131 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 141 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 151 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 $1^{st}$ 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 161 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 $3^{rd}$ 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 $3^{rd}$ quadarant] $=\tan\theta$

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