True False[1 Marks ]

Question 11 Mark

If $\tan\theta+\tan2\theta+\sqrt{3}\tan\theta\tan2\theta=\sqrt{3},$ then $\theta=\frac{\text{n}\pi}{3}+\frac{\pi}{9}$

Answer

True.
Solution:
Given that, $\tan\theta+\tan2\theta=-\sqrt3\tan\theta\tan2\theta+\sqrt3$
$\Rightarrow\tan\theta+\tan2\theta=\sqrt3-\sqrt3\tan\theta\tan2\theta$
$\Rightarrow\tan\theta+\tan2\theta=\sqrt3(1-\tan\theta\tan2\theta)$
$\Rightarrow\frac{\tan\theta+\tan2\theta}{1-\tan\theta\tan2\theta}=\sqrt3$
$\Rightarrow\tan(\theta+2\theta)=\sqrt3$
$\Rightarrow\tan3\theta=\tan\frac{\pi}{3}\therefore3\theta=\text{n}\pi+\frac{\pi}{3}$
so $\theta=\frac{\text{n}\pi}{3}+\frac{\pi}{9}$
Hence, the given statement is 'true'.

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Question 21 Mark

$\sin10^\circ$ is greater than $\cos10^\circ.$

Answer

False.
Solution:
$$If $\sin10^\circ>\cos10^\circ$
$\Rightarrow\sin10^\circ>\cos(90^\circ-80^\circ)$
$\Rightarrow\sin10^\circ>\sin80^\circ$ Which is not possible, because value of sine is in increasing order .
Hence, the statement is 'False'.

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Question 31 Mark

One value of $\theta$ which satisfies the equation $\sin^4\theta-2\sin^2\theta-1$ lies between 0 and $2\pi.$

Answer

False'
Solution:
Given equation is
$\sin^4\theta-2\sin^2\theta=0$
$\sin^2\theta=\frac{-(-2)\pm\sqrt{(-2)-4\times1\times(-1)}}{2\times1}$
$=\frac{2\pm\sqrt{4+4}}{2}$
$=\frac{2\pm\sqrt8}{2}$
$=\frac{2\pm2\sqrt2}{2}$
$=1\pm\sqrt2$
$\therefore\sin^2=(1+\sqrt2)$ or $(1-\sqrt2)$
$\Rightarrow-1\leq\sin\theta\leq1$
$\Rightarrow\sin^2\theta\leq1$but $\sin^2\theta=(1+\sqrt2)$ or $(1-\sqrt2)$
which is not possible.
Hence, the given statement is 'False'

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Question 41 Mark

If $\tan\text{A}=\frac{1-\cos\text{B}}{\sin\text{B}},$ then $\tan2\text{A}=\tan\text{B}$

Answer

True.
Solution:
Given that, $\tan\text{A}=\frac{1-\cos\text{B}}{\sin\text{B}}=\frac{2\sin^2\frac{\text{B}}{2}}{2\sin\frac{\text{B}}{2}2\cos\frac{\text{B}}{2}}=\tan\frac{\text{B}}{2}$
$\tan2\text{A}=\frac{2\tan\text{A}}{1-\tan^2\text{A}}=\frac{2\tan\frac{\text{B}}{2}}{1-\tan^2\frac{\text{B}}{2}}$
$\therefore2\text{A}=\tan\text{B}$
Hence, the statement is 'True'.

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Question 51 Mark

The equality $\sin\text{A}+\sin2\text{A}+\sin3\text{A}=3$ holds for some real value of A.

Answer

False.
Solution:
Given that, $\sin\text{A}+\sin2\text{A}+\sin3\text{A}=3$
since the maximum value of $\sin\text{A}$ is 1 but for $\sin2\text{A}$ and $\sin3\text{A}$ it is not equal to 1. so it is not possible.
Hence, the given statement is 'False'.

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Question 61 Mark

$\cos\frac{2\pi}{15}\cos\frac{4\pi}{15}\cos\frac{8\pi}{15}\cos\frac{16\pi}{15}=\frac{1}{16}$

Answer

True.
Solution:
$\text{L.H.S.}\cos\frac{2\pi}{15}\cdot\cos\frac{4\pi}{15}\cdot\cos\frac{8\pi}{15}\cdot\cos\frac{16\pi}{15}$
$=\cos24^\circ.\cos48^\circ.\cos96^\circ.\cos192^\circ$
$=\frac{1}{16\sin24^\circ}[2\sin24^\circ\cos24^\circ)(2\cos48^\circ)(2\cos96^\circ)(2\cos192^\circ)]$
$=\frac{1}{16\sin24^\circ}[\sin48^\circ\cdot2\cos48^\circ(2\cos96^\circ)(2\cos192^\circ)]$
$=\frac{1}{16\sin24^\circ}[2\sin48^\circ\cos48^\circ(2\cos96^\circ)(2\cos192^\circ)]$
$\frac{1}{16\sin24^\circ}[\sin96^\circ(2\cos96^\circ)(2\cos192^\circ)]$
$=\frac{1}{16\sin24^\circ}[2\sin96^\circ\cdot\cos96^\circ(2\cos192^\circ)]$
$=\frac{1}{16\sin24^\circ}[\sin192^\circ\cdot(2\cos192^\circ)]$
$=\frac{1}{16\sin24^\circ}2\sin192^\circ\cos192^\circ$
$=\frac{1}{16\sin24^\circ}\sin384^\circ=\frac{1}{16\sin24^\circ}\sin(360^\circ+24^\circ)$
$=\frac{1}{16\sin24^\circ}\times\sin24^\circ[\because\sin(360^\circ+\theta)=\sin\theta]$
$=\frac{1}{16}\text{R.H.S.}$
Hence, the given statement is 'True'.

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Question 71 Mark

If $\tan(\pi\cos\theta)=\cot(\pi\sin\theta),$ then $\cos\Big(\theta-\frac{\pi}{4}\Big)=\pm\frac{1}{2\sqrt{2}}$

Answer

True.
Solution:
Given that, $\tan(\pi\cos\theta)=\cot(\pi\sin\theta)$
$\Rightarrow\tan(\pi\cos\theta)=\tan\Big(\frac{\pi}{2}-\pi\sin\theta\Big)$
$\Rightarrow\pi\cos\theta=\frac{\pi}{2}-\pi\sin\theta$
$\Rightarrow\pi\cos\theta+\pi\sin\theta=\frac{\pi}{2}$
$\Rightarrow\cos\theta+\sin\theta=\frac{1}{2}$
$\Rightarrow\frac{1}{\sqrt2}\cos\theta+\frac{1}{\sqrt2}\sin\theta=\frac{1}{2\sqrt2}$
$\Rightarrow\cos\frac{\pi}{4}\cos\theta+\sin\frac{\pi}{4}\sin\theta=\frac{1}{2\sqrt2}$$$
$\Rightarrow\cos\Big(\theta-\frac{\pi}{4}\Big)=\pm\frac{1}{2\sqrt2}$$\Big[\because\cos\Big(\theta-\frac{\pi}{4}\Big)$or $\cos\Big(\frac{\pi}{4}-\theta\Big)\Big]$
Hence, the given statement is 'true'.

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Question 81 Mark

If $\text{cosec x }=1+\cot\text{x}$ then $\text{x}=2\text{n}\pi,2\text{n}\pi+\frac{\pi}{2}$

Answer

True.
Solution:
Given that, $\text{cosec}\text{x}=1+\cot\text{x}$
$\Rightarrow\frac{1}{\sin\text{x}}=1+\frac{\cos\text{x}}{\sin\text{x}}$
$\Rightarrow\frac{1}{\sin\text{x}}=\frac{\sin\text{x}+\cos\text{x}}{\sin\text{x}}$
$\Rightarrow\sin\text{x}+\cos\text{x}=1$
$\Rightarrow\frac{1}{\sqrt2}\sin\text{x}+\frac{1}{\sqrt{2}}\cos\text{x}=\frac{1}{\sqrt2}$
$\Rightarrow\sin\frac{\pi}{4}\sin\text{x}+\cos\frac{\pi}{4}\cos\text{x}=\frac{1}{\sqrt2}$
$\Rightarrow\cos\Big(\text{x}-\frac{\pi}{4}\Big)=\frac{1}{\sqrt2}$
$\Rightarrow\cos\Big(\text{x}-\frac{\pi}{4}\Big)=\cos\frac{\pi}{4}$
$\text{x}=2\text{n}\pi+\frac{\pi}{4}+\frac{\pi}{4}\Rightarrow\text{x}=2\text{n}\pi+\frac{\pi}{2}$
or $\text{x}=2\text{n}\pi+\frac{\pi}{4}-\frac{\pi}{4}\Rightarrow\text{x}=2\text{n}\pi$
Hence, the given statement is 'True'.

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Maths True False[1 Marks ]

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