Question

Find the exact value of:
(a) $ta{n}^2\frac{\pi }{16}+ta{n}^2\frac{2\pi }{16}+ta{n}^2\frac{3\pi }{16}+......+ta{n}^2\frac{7\pi }{16}$
(b) ${\left(\frac{2+sec{20}^o}{cosec{20}^o}\right)}^2$

Answer 1

(i) Given ${\tan }^2\frac{\pi }{16}+{\tan }^2\frac{2\pi }{16}+.......+{\tan }^2\frac{7\pi }{16}$

$\Rightarrow$ We know that $\tan \theta =\cot (\pi -\theta )$

By making Paris we get

$\Rightarrow ({\tan }^2\frac{\pi }{16}+{\cot }^2\frac{\pi }{16})+({\tan }^2\frac{2\pi }{16}+{\cot }^2\frac{2\pi }{16})+({\tan }^2\frac{3\pi }{16}+{\cot }^2\frac{3\pi }{16})+{\tan }^2\frac{\pi }{4}$

$\Rightarrow$ using identify ${\tan }^{2}x+{\cot }^{2}x=4{\csc }^{2}2x-2$ we get

$\Rightarrow 4({\csc }^2\frac{\pi }{8}+{\csc }^2\frac{\pi }{2}+{\csc }^2\frac{3\pi }{8})-6+1$

$\Rightarrow 4({\csc }^2\frac{\pi }{8}+{\csc }^2\frac{3\pi }{8})+4-5$

now we know $\csc \frac{3\pi }{8}=\csc (\frac{\pi }{2}-\frac{\pi }{8})=\sec \frac{\pi }{8}$

$\Rightarrow 4\left(\frac{{\sin }^2\frac{\pi }{8}+{\cos }^2\frac{\pi }{8}}{{\sin }^2\frac{\pi }{8}{\cos }^2\frac{\pi }{8}}\right)-1$

$\Rightarrow \frac{4\times 4}{{\sin }^2\frac{\pi }{4}}-1\Rightarrow \frac{16}{(1/\sqrt{2}{)}^2}-1=32-1$

$[\Rightarrow 31]dx$

(ii) ${\left(\frac{2+\sec 20}{\csc 20}\right)}^2\Rightarrow {\left(\frac{2+\frac{1}{\cos 20}}{\frac{1}{\sin 20}}\right)}^2$

$\Rightarrow \left(\frac{2\cos 20\sin 20+\sin 20}{\cos 20}\right)\Rightarrow {\left(\frac{\sin 40+\sin 20}{\cos 20}\right)}^2$

$\Rightarrow$ we know $\sin a+\sin b=2\sin \left(\frac{a+b}{2}\right)\cos \left(\frac{a-b}{2}\right)$

$\Rightarrow \left(\frac{2\sin 30\cos 20}{\cos 20}\right)\Rightarrow (2\sin 30{)}^2$

$\Rightarrow (2\times \frac{1}{2})=1$