Question

$tan[\frac{\pi }{4}+\frac{1}{2}co{s}^{-1}\frac{a}{b}]+tan[\frac{\pi }{4}+\frac{1}{2}co{s}^{-1}\frac{a}{b}]$
[MP PET 1999]

• A.
• B.
• C.
• D.

Answer 1

The correct option is B

tan $[\frac{\pi }{4}+\frac{1}{2}co{s}^{-1}\frac{a}{b}]+tan[\frac{\pi }{4}+\frac{1}{2}co{s}^{-1}\frac{a}{b}]Let\frac{1}{2}co{s}^{-1}\frac{a}{b}=\theta \Rightarrow cos2\theta =\frac{a}{b}Thus,tan[\frac{\pi }{4}+\theta ]+tan[\frac{\pi }{4}-\theta ]=\frac{1+tan\theta }{1-tan\theta }+\frac{1-tan\theta }{1+tan\theta }=\frac{(1+tan\theta {)}^2+(1-tan\theta {)}^2}{(1-ta{n}^2\theta )}\frac{1+ta{n}^2\theta +2tan\theta +1+ta{n}^2\theta -2tan\theta }{(1+ta{n}^2\theta )}=\frac{2(1+ta{n}^2\theta )}{1-ta{n}^2\theta }=2sec2\theta =\frac{2}{cos2\theta }=\frac{2}{a/b}=\frac{2b}{a}$