Class 11: Transformation Formulae – Very Short Answer Questions – ICSE / ISC / CBSE Mathematics Portal for K12 Students

$\displaystyle \textbf{Question 1. }\text{If }(\cos\alpha+\cos\beta)^2+(\sin\alpha+\sin\beta)^2=\lambda\cos^2\left(\frac{\alpha-\beta}{2}\right),\text{ write the value of }\lambda.$
$\displaystyle \text{Answer:}$
$\displaystyle (\cos\alpha+\cos\beta)^2+(\sin\alpha+\sin\beta)^2$
$\displaystyle =2+2(\cos\alpha\cos\beta+\sin\alpha\sin\beta)$
$\displaystyle =2+2\cos(\alpha-\beta)$
$\displaystyle =4\cos^2\left(\frac{\alpha-\beta}{2}\right)$
$\displaystyle \therefore \lambda=4$
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$\displaystyle \textbf{Question 2. }\text{Write the value of }\sin\frac{\pi}{12}\sin\frac{5\pi}{12}.$
$\displaystyle \text{Answer:}$
$\displaystyle \sin\frac{\pi}{12}\sin\frac{5\pi}{12}=\sin15^\circ\sin75^\circ$
$\displaystyle =\frac{\cos60^\circ-\cos90^\circ}{2}$
$\displaystyle =\frac{1}{4}$
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$\displaystyle \textbf{Question 3. }\text{If }\sin A+\sin B=\alpha\text{ and }\cos A+\cos B=\beta,\text{ then write the value of }\tan\left(\frac{A+B}{2}\right).$
$\displaystyle \text{Answer:}$
$\displaystyle \sin A+\sin B=2\sin\frac{A+B}{2}\cos\frac{A-B}{2}=\alpha$
$\displaystyle \cos A+\cos B=2\cos\frac{A+B}{2}\cos\frac{A-B}{2}=\beta$
$\displaystyle \therefore \tan\left(\frac{A+B}{2}\right)=\frac{\alpha}{\beta}$
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$\displaystyle \textbf{Question 4. }\text{If }\cos A=m\cos B,\text{ then write the value of }\cot\frac{A+B}{2}\cot\frac{A-B}{2}.$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{\cos A+\cos B}{\cos A-\cos B}=\frac{m+1}{m-1}$
$\displaystyle \frac{2\cos\frac{A+B}{2}\cos\frac{A-B}{2}}{-2\sin\frac{A+B}{2}\sin\frac{A-B}{2}}=\frac{m+1}{m-1}$
$\displaystyle -\cot\frac{A+B}{2}\cot\frac{A-B}{2}=\frac{m+1}{m-1}$
$\displaystyle \therefore \cot\frac{A+B}{2}\cot\frac{A-B}{2}=-\frac{m+1}{m-1}$
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$\displaystyle \textbf{Question 5. }\text{Write the value of the expression }\frac{1-4\sin10^\circ\sin70^\circ}{2\sin10^\circ}.$
$\displaystyle \text{Answer:}$
$\displaystyle 4\sin10^\circ\sin70^\circ=2(\cos60^\circ-\cos80^\circ)$
$\displaystyle =1-2\cos80^\circ$
$\displaystyle 1-4\sin10^\circ\sin70^\circ=2\cos80^\circ$
$\displaystyle \therefore \frac{1-4\sin10^\circ\sin70^\circ}{2\sin10^\circ}=\frac{2\cos80^\circ}{2\sin10^\circ}$
$\displaystyle =1$
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$\displaystyle \textbf{Question 6. }\text{If }A+B=\frac{\pi}{3}\text{ and }\cos A+\cos B=1,\text{ then find the value of }\cos\frac{A-B}{2}.$
$\displaystyle \text{Answer:}$
$\displaystyle \cos A+\cos B=2\cos\frac{A+B}{2}\cos\frac{A-B}{2}$
$\displaystyle 1=2\cos\frac{\pi}{6}\cos\frac{A-B}{2}$
$\displaystyle 1=\sqrt3\cos\frac{A-B}{2}$
$\displaystyle \therefore \cos\frac{A-B}{2}=\frac{1}{\sqrt3}$
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$\displaystyle \textbf{Question 7. }\text{Write the value of }\sin12^\circ\sin48^\circ\sin54^\circ.$
$\displaystyle \text{Answer:}$
$\displaystyle \sin54^\circ=\cos36^\circ$
$\displaystyle \sin12^\circ\sin48^\circ=\frac{\cos36^\circ-\cos60^\circ}{2}$
$\displaystyle \therefore \sin12^\circ\sin48^\circ\sin54^\circ$
$\displaystyle =\frac{\cos36^\circ-\frac{1}{2}}{2}\cos36^\circ$
$\displaystyle =\frac{1}{8}$
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$\displaystyle \textbf{Question 8. }\text{If }\sin2A=\lambda\sin2B,\text{ then write the value of }\frac{\lambda+1}{\lambda-1}.$
$\displaystyle \text{Answer:}$
$\displaystyle \lambda=\frac{\sin2A}{\sin2B}$
$\displaystyle \frac{\lambda+1}{\lambda-1}=\frac{\sin2A+\sin2B}{\sin2A-\sin2B}$
$\displaystyle =\frac{2\sin(A+B)\cos(A-B)}{2\cos(A+B)\sin(A-B)}$
$\displaystyle =\tan(A+B)\cot(A-B)$
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$\displaystyle \textbf{Question 9. }\text{Write the value of }\frac{\sin A+\sin3A}{\cos A+\cos3A}.$
$\displaystyle \text{Answer:}$
$\displaystyle \sin A+\sin3A=2\sin2A\cos A$
$\displaystyle \cos A+\cos3A=2\cos2A\cos A$
$\displaystyle \therefore \frac{\sin A+\sin3A}{\cos A+\cos3A}=\tan2A$
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$\displaystyle \textbf{Question 10. }\text{If }\cos(A+B)\sin(C-D)=\cos(A-B)\sin(C+D),\text{ then write the value of} \\ \tan A\tan B\tan C.$
$\displaystyle \text{Answer:}$
$\displaystyle \cos(A+B)(\sin C\cos D-\cos C\sin D)=\cos(A-B)(\sin C\cos D+\cos C\sin D)$
$\displaystyle \{\cos(A+B)-\cos(A-B)\}\sin C\cos D$
$\displaystyle =\{\cos(A+B)+\cos(A-B)\}\cos C\sin D$
$\displaystyle -2\sin A\sin B\sin C\cos D=2\cos A\cos B\cos C\sin D$
$\displaystyle \therefore \tan A\tan B\tan C=-\tan D$
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