\displaystyle \textbf{Question 1. }\text{If }\cos4x=1+k\sin^2x\cos^2x,\text{ then write the value of }k.
\displaystyle \text{Answer:}
\displaystyle \cos4x=1-2\sin^22x
\displaystyle =1-2(2\sin x\cos x)^2
\displaystyle =1-8\sin^2x\cos^2x
\displaystyle \therefore k=-8
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\displaystyle \textbf{Question 2. }\text{If }\tan\frac{x}{2}=\frac{m}{n},\text{ then write the value of }m\sin x+n\cos x.
\displaystyle \text{Answer:}
\displaystyle \sin x=\frac{2mn}{m^2+n^2},\quad \cos x=\frac{n^2-m^2}{m^2+n^2}
\displaystyle m\sin x+n\cos x=\frac{2m^2n+n^3-nm^2}{m^2+n^2}
\displaystyle =\frac{n(m^2+n^2)}{m^2+n^2}=n
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\displaystyle \textbf{Question 3. }\text{If }\frac{\pi}{2}<\theta<\frac{3\pi}{2},\text{ then write the value of }\sqrt{\frac{1+\cos2\theta}{2}}.
\displaystyle \text{Answer:}
\displaystyle \sqrt{\frac{1+\cos2\theta}{2}}=\sqrt{\cos^2\theta}=|\cos\theta|
\displaystyle \frac{\pi}{2}<\theta<\frac{3\pi}{2}\Rightarrow \cos\theta<0
\displaystyle \therefore |\cos\theta|=-\cos\theta
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\displaystyle \textbf{Question 4. }\text{If }\frac{\pi}{2}<\theta<\pi,\text{ then write the value of }\sqrt{2+\sqrt{2+2\cos2\theta}}\text{ in the simplest form.}
\displaystyle \text{Answer:}
\displaystyle \sqrt{2+2\cos2\theta}=\sqrt{4\cos^2\theta}=2|\cos\theta|
\displaystyle \frac{\pi}{2}<\theta<\pi\Rightarrow |\cos\theta|=-\cos\theta
\displaystyle \therefore \sqrt{2+\sqrt{2+2\cos2\theta}}=\sqrt{2-2\cos\theta}
\displaystyle =2\sin\frac{\theta}{2}
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\displaystyle \textbf{Question 5. }\text{If }\frac{\pi}{2}<\theta<\pi,\text{ then write the value of }\sqrt{\frac{1-\cos2\theta}{1+\cos2\theta}}.
\displaystyle \text{Answer:}
\displaystyle \sqrt{\frac{1-\cos2\theta}{1+\cos2\theta}}=\sqrt{\frac{2\sin^2\theta}{2\cos^2\theta}}
\displaystyle =|\tan\theta|
\displaystyle \frac{\pi}{2}<\theta<\pi\Rightarrow \tan\theta<0
\displaystyle \therefore |\tan\theta|=-\tan\theta
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\displaystyle \textbf{Question 6. }\text{If }\pi<\theta<\frac{3\pi}{2},\text{ then write the value of }\sqrt{\frac{1-\cos2\theta}{1+\cos2\theta}}.
\displaystyle \text{Answer:}
\displaystyle \sqrt{\frac{1-\cos2\theta}{1+\cos2\theta}}=|\tan\theta|
\displaystyle \pi<\theta<\frac{3\pi}{2}\Rightarrow \tan\theta>0
\displaystyle \therefore |\tan\theta|=\tan\theta
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\displaystyle \textbf{Question 7. }\text{In a right angled triangle }ABC,\text{ write the value of }\sin^2A+\sin^2B+\sin^2C.
\displaystyle \text{Answer:}
\displaystyle \text{Let }C=90^\circ,\text{ then }A+B=90^\circ
\displaystyle \sin^2A+\sin^2B+\sin^2C=\sin^2A+\cos^2A+1
\displaystyle =2
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\displaystyle \textbf{Question 8. }\text{Write the value of }\cos^276^\circ+\cos^216^\circ-\cos76^\circ\cos16^\circ.
\displaystyle \text{Answer:}
\displaystyle \cos76^\circ=\sin14^\circ
\displaystyle \therefore S=\sin^2 14^\circ+\cos^2 16^\circ-\sin14^\circ\cos16^\circ
\displaystyle \cos16^\circ=\sin74^\circ
\displaystyle S=\sin^2 14^\circ+\sin^2 74^\circ-\sin14^\circ\sin74^\circ
\displaystyle \sin74^\circ=\cos16^\circ
\displaystyle \sin14^\circ=\cos76^\circ
\displaystyle \text{Using }a^2+b^2-ab=\frac{1}{2}\left[(a-b)^2+a^2+b^2\right]
\displaystyle \text{or directly using values,}
\displaystyle S=\frac{3}{4}
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\displaystyle \textbf{Question 9. }\text{If }\frac{\pi}{4}<\theta<\frac{\pi}{2},\text{ then write the value of }\sqrt{1-\sin2\theta}.
\displaystyle \text{Answer:}
\displaystyle 1-\sin2\theta=(\sin\theta-\cos\theta)^2
\displaystyle \sqrt{1-\sin2\theta}=|\sin\theta-\cos\theta|
\displaystyle \frac{\pi}{4}<\theta<\frac{\pi}{2}\Rightarrow \sin\theta>\cos\theta
\displaystyle \therefore \sqrt{1-\sin2\theta}=\sin\theta-\cos\theta
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\displaystyle \textbf{Question 10. }\text{Write the value of }\cos\frac{\pi}{7}\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}.
\displaystyle \text{Answer:}
\displaystyle \cos\frac{\pi}{7}\cos\frac{2\pi}{7}\cos\frac{4\pi}{7}=-\frac{1}{8}
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\displaystyle \textbf{Question 11. }\text{If }\tan A=\frac{1-\cos B}{\sin B},\text{ then find the value of }\tan2A.
\displaystyle \text{Answer:}
\displaystyle \frac{1-\cos B}{\sin B}=\tan\frac{B}{2}
\displaystyle \therefore \tan A=\tan\frac{B}{2}
\displaystyle \therefore A=\frac{B}{2}
\displaystyle \therefore \tan2A=\tan B
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\displaystyle \textbf{Question 12. }\text{If }\sin x+\cos x=a,\text{ find the value of }\sin^6x+\cos^6x.
\displaystyle \text{Answer:}
\displaystyle (\sin x+\cos x)^2=a^2
\displaystyle 1+2\sin x\cos x=a^2
\displaystyle \sin x\cos x=\frac{a^2-1}{2}
\displaystyle \sin^6x+\cos^6x=(\sin^2x+\cos^2x)^3-3\sin^2x\cos^2x(\sin^2x+\cos^2x)
\displaystyle =1-3\left(\frac{a^2-1}{2}\right)^2
\displaystyle =1-\frac{3}{4}(a^2-1)^2
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\displaystyle \textbf{Question 13. }\text{If }\sin x+\cos x=a,\text{ find the value of }|\sin x-\cos x|.
\displaystyle \text{Answer:}
\displaystyle (\sin x+\cos x)^2+(\sin x-\cos x)^2=2
\displaystyle a^2+(\sin x-\cos x)^2=2
\displaystyle |\sin x-\cos x|=\sqrt{2-a^2}
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