\displaystyle \textbf{Question 1. }\text{If }\alpha+\beta-\gamma=\pi,\text{ and }\sin^2\alpha+\sin^2\beta-\sin^2\gamma=\lambda\sin\alpha\sin\beta\cos\gamma, \\ \text{ then write the value of }\lambda.
\displaystyle \text{Answer:}
\displaystyle \alpha+\beta-\gamma=\pi
\displaystyle \therefore \gamma=\alpha+\beta-\pi
\displaystyle \sin\gamma=\sin(\alpha+\beta-\pi)=-\sin(\alpha+\beta)
\displaystyle \therefore \sin^2\gamma=\sin^2(\alpha+\beta)
\displaystyle \sin^2\alpha+\sin^2\beta-\sin^2(\alpha+\beta)
\displaystyle =-2\sin\alpha\sin\beta\cos(\alpha+\beta)
\displaystyle \cos\gamma=\cos(\alpha+\beta-\pi)=-\cos(\alpha+\beta)
\displaystyle \therefore -2\sin\alpha\sin\beta\cos(\alpha+\beta)=2\sin\alpha\sin\beta\cos\gamma
\displaystyle \therefore \lambda=2
\\

\displaystyle \textbf{Question 2. }\text{If }x\cos\theta=y\cos\left(\theta+\frac{2\pi}{3}\right)=z\cos\left(\theta+\frac{4\pi}{3}\right),\text{ then write the value of } \\ \frac{1}{x}+\frac{1}{y}+\frac{1}{z}.
\displaystyle \text{Answer:}
\displaystyle \text{Let }x\cos\theta=y\cos\left(\theta+\frac{2\pi}{3}\right)=z\cos\left(\theta+\frac{4\pi}{3}\right)=k
\displaystyle \therefore \frac{1}{x}=\frac{\cos\theta}{k},\quad \frac{1}{y}=\frac{\cos\left(\theta+\frac{2\pi}{3}\right)}{k},\quad \frac{1}{z}=\frac{\cos\left(\theta+\frac{4\pi}{3}\right)}{k}
\displaystyle \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{k}\left\{\cos\theta+\cos\left(\theta+\frac{2\pi}{3}\right)+\cos\left(\theta+\frac{4\pi}{3}\right)\right\}
\displaystyle \cos\theta+\cos\left(\theta+\frac{2\pi}{3}\right)+\cos\left(\theta+\frac{4\pi}{3}\right)=0
\displaystyle \therefore \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0
\\

\displaystyle \textbf{Question 3. }\text{Write the maximum and minimum values of } \\ 3\cos x+4\sin x+5.
\displaystyle \text{Answer:}
\displaystyle 3\cos x+4\sin x\in[-5,5]
\displaystyle \therefore 3\cos x+4\sin x+5\in[0,10]
\displaystyle \text{Maximum value}=10
\displaystyle \text{Minimum value}=0
\\

\displaystyle \textbf{Question 4. }\text{Write the maximum value of }12\sin\theta-9\sin^2\theta.
\displaystyle \text{Answer:}
\displaystyle \text{Let }\sin\theta=t,\text{ where }-1\leq t\leq1
\displaystyle 12\sin\theta-9\sin^2\theta=12t-9t^2
\displaystyle =-9\left(t-\frac{2}{3}\right)^2+4
\displaystyle \therefore \text{Maximum value}=4
\\

\displaystyle \textbf{Question 5. }\text{If }12\sin\theta-9\sin^2\theta\text{ attains its maximum value at } \\ \theta=\alpha,\text{ then write the value of }\sin\alpha.
\displaystyle \text{Answer:}
\displaystyle 12\sin\theta-9\sin^2\theta=-9\left(\sin\theta-\frac{2}{3}\right)^2+4
\displaystyle \text{Maximum value occurs when }\sin\theta=\frac{2}{3}
\displaystyle \therefore \sin\alpha=\frac{2}{3}
\\

\displaystyle \textbf{Question 6. }\text{Write the interval in which the values of } \\ 5\cos\theta+3\cos\left(\theta+\frac{\pi}{3}\right)+3\text{ lie.}
\displaystyle \text{Answer:}
\displaystyle 5\cos\theta+3\cos\left(\theta+\frac{\pi}{3}\right)+3
\displaystyle =5\cos\theta+3\left(\frac{1}{2}\cos\theta-\frac{\sqrt3}{2}\sin\theta\right)+3
\displaystyle =\frac{13}{2}\cos\theta-\frac{3\sqrt3}{2}\sin\theta+3
\displaystyle \text{Amplitude}=\sqrt{\left(\frac{13}{2}\right)^2+\left(\frac{3\sqrt3}{2}\right)^2}=\sqrt{49}=7
\displaystyle \therefore \text{Required interval}=[3-7,3+7]=[-4,10]
\\

\displaystyle \textbf{Question 7. }\text{If }\tan(A+B)=p\text{ and }\tan(A-B)=q,\text{ then write the value of }\tan2B.
\displaystyle \text{Answer:}
\displaystyle 2B=(A+B)-(A-B)
\displaystyle \tan2B=\frac{\tan(A+B)-\tan(A-B)}{1+\tan(A+B)\tan(A-B)}
\displaystyle \therefore \tan2B=\frac{p-q}{1+pq}
\\

\displaystyle \textbf{Question 8. }\text{If }\frac{\cos(x-y)}{\cos(x+y)}=\frac{m}{n},\text{ then write the value of }\tan x\tan y.
\displaystyle \text{Answer:}
\displaystyle \frac{\cos(x-y)}{\cos(x+y)}=\frac{\cos x\cos y+\sin x\sin y}{\cos x\cos y-\sin x\sin y}
\displaystyle =\frac{1+\tan x\tan y}{1-\tan x\tan y}=\frac{m}{n}
\displaystyle n(1+\tan x\tan y)=m(1-\tan x\tan y)
\displaystyle (m+n)\tan x\tan y=m-n
\displaystyle \therefore \tan x\tan y=\frac{m-n}{m+n}
\\

\displaystyle \textbf{Question 9. }\text{If }a=b\cos\frac{2\pi}{3}=c\cos\frac{4\pi}{3},\text{ then write the value of }ab+bc+ca.
\displaystyle \text{Answer:}
\displaystyle \cos\frac{2\pi}{3}=\cos\frac{4\pi}{3}=-\frac{1}{2}
\displaystyle a=-\frac{b}{2}=-\frac{c}{2}
\displaystyle \therefore b=-2a,\quad c=-2a
\displaystyle ab+bc+ca=a(-2a)+(-2a)(-2a)+(-2a)a
\displaystyle =-2a^2+4a^2-2a^2=0
\\

\displaystyle \textbf{Question 10. }\text{If }A+B=C,\text{ then write the value of }\tan A\tan B\tan C.
\displaystyle \text{Answer:}
\displaystyle A+B=C
\displaystyle \tan(A+B)=\tan C
\displaystyle \frac{\tan A+\tan B}{1-\tan A\tan B}=\tan C
\displaystyle \tan A+\tan B=\tan C-\tan A\tan B\tan C
\displaystyle \therefore\tan A\tan B\tan C=\tan C-\tan A-\tan B
\\

\displaystyle \textbf{Question 11. }\text{If }\sin\alpha-\sin\beta=a\text{ and }\cos\alpha+\cos\beta=b,\text{ then write the value of } \\ \cos(\alpha+\beta).
\displaystyle \text{Answer:}
\displaystyle \sin\alpha-\sin\beta=2\cos\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}=a
\displaystyle \cos\alpha+\cos\beta=2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}=b
\displaystyle a^2+b^2=4\cos^2\frac{\alpha+\beta}{2}
\displaystyle \cos(\alpha+\beta)=2\cos^2\frac{\alpha+\beta}{2}-1
\displaystyle \therefore \cos(\alpha+\beta)=\frac{a^2+b^2}{2}-1
\\

\displaystyle \textbf{Question 12. }\text{If }\tan\alpha=\frac{1}{1+2^{-x}}\text{ and }\tan\beta=\frac{1}{1+2^{x+1}},\text{ then write the value of } \\ \alpha+\beta\text{ lying in the interval }(0,\pi/2).
\displaystyle \text{Answer:}
\displaystyle \text{Let }2^x=t
\displaystyle \tan\alpha=\frac{t}{t+1},\quad \tan\beta=\frac{1}{1+2t}
\displaystyle \tan(\alpha+\beta)=\frac{\frac{t}{t+1}+\frac{1}{1+2t}}{1-\frac{t}{(t+1)(1+2t)}}
\displaystyle =1
\displaystyle \alpha+\beta\in\left(0,\frac{\pi}{2}\right)
\displaystyle \therefore \alpha+\beta=\frac{\pi}{4}
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