2 Marks Questions

Question 12 Marks

If $\sin\theta+\cos\theta=1,$ then find the general value of $\theta$

Answer

$\sin\theta+\cos\theta=1$
$\Rightarrow\frac{1}{\sqrt2}\sin\theta+\frac{1}{\sqrt2}\cos\theta=\frac{1}{\sqrt2}\Rightarrow\sin\theta\cos\frac{\pi}{4}+\cos\theta\sin\frac{\pi}{4}=\frac{1}{\sqrt2}$
$\Rightarrow\sin\Big(\theta+\frac{\pi}{4}\Big)=\sin\frac{\pi}{4}\Rightarrow\theta+\frac{\pi}{4}=\text{n}\pi+(-1)^\text{n}\frac{\pi}{4},\text{n}\in\text{Z}$
$\therefore\theta=\text{n}\pi(-1)^\text{n}\frac{\pi}{4}-\frac{\pi}{4},\text{n}\in\text{Z}$

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Question 22 Marks

If $\tan(\text{A + B})=\text{p},\tan(\text{A}-\text{B})=\text{q},$ then show that $\tan2\text{A}=\frac{\text{p + q}}{1-\text{pq}}$
$\big[$Hint: Use $2\text{A}=(\text{A + B})+(\text{A}-\text{B})\big]$

Answer

We have $\tan(\text{A + B})=\text{p}$ and $\tan(\text{A}-\text{B})=\text{q}$
$\tan2\text{A}=\tan[(\text{A + B})-(\text{A}-\text{B})]$
$=\frac{\tan(\text{A + B})+\tan(\text{A}-\text{B})}{1-\tan(\text{A + B})\tan(\text{A}-\text{B})}=\frac{\text{p + q}}{1-\text{pq}}$

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Question 32 Marks

Prove that $4\text{A}=4\sin\text{A}\cos^3\text{A}-4\cos\text{A}\sin^3\text{A}.$

Answer

$\text{L.H.S.}=\sin4\text{A}$
$=2\sin2\text{A}-\cos2\text{A}=2(2\sin\text{A}\cos\text{A})(\cos^2\text{A}-\sin^2\text{A})$
$=4\sin\text{A}.\cos^3\text{A}-4\cos\text{A}\sin^3\text{A}=\text{R.H.S.}$

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Question 42 Marks

If $2\sin^2\theta=3\cos\theta,$ where $0\leq\theta\leq2\pi,$ then find the value of $\theta.$

Answer

We have, $2\sin^2\theta=3\cos\theta $
$\Rightarrow2-2\cos^2\theta=3\cos\theta$
$\Rightarrow2\cos^2\theta+3\cos\theta-2=0$
$\Rightarrow(2\cos\theta+2)(2\cos\theta-1)=0$
$\therefore\cos\theta=\frac{1}{2}=\cos\frac{\pi}{3}$
$\therefore\theta=\frac{\pi}{3}$ Or $2\pi-\frac{\pi}{3}$
$\therefore\theta=\frac{\pi}{3},\frac{5\pi}{3}$

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Question 52 Marks

Find the value of $\tan22^{\circ}30'.$

Answer

We know that, $\tan\frac{\theta}{2}=\frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}}=\frac{2\sin\frac{\theta}{2}.\cos\frac{\theta}{2}}{2\cos^2\frac{\theta}{2}}=\frac{\sin\theta}{1+\cos\theta}$
$\therefore\ \tan22^{\circ}30'=\frac{\sin45^{\circ}}{1+\cos45^{\circ}}=\frac{\frac{1}{\sqrt2}}{1+\frac{1}{\sqrt2}}=\frac{1}{\sqrt2+1}$

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Question 62 Marks

Find the most general value of $\theta$ satisfying the equation $\tan\theta=-1$ and $\cos\theta=\frac{1}{\sqrt{2}}.$

Answer

we have $\tan\theta=-1$ and $\cos\theta=\frac{1}{\sqrt2}.$
so, $\theta$ lies in IV quadrant.
$\theta=\frac{7}{4}$
so general solution is $\theta=\frac{7\pi}{4}+2\text{n}\pi,\text{n}\in\text{Z}$$$

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