Question

# If $$\displaystyle e^{\displaystyle (\sin^2 x+\sin^4 x+\sin^6 x + ..... \infty) \ln 3}$$, $$\displaystyle x \epsilon (0, \frac {\pi}{2})$$ satisfies the equation $$\displaystyle t^2 - 28t + 27 = 0$$ then value of $$\displaystyle (\cos \: x + \sin \: x)^{-1}$$ equals

A
31
B
2(31)
C
3+1
D
1

Solution

## The correct option is A $$\displaystyle \sqrt 3 - 1$$\sin ce $$\displaystyle \sin ^2x < 1$$ for $$x\epsilon (0, \pi/2)$$ So, $$\displaystyle e^{ (\sin ^2x + \sin ^4x + \sin ^6x + .......\infty)\ln3}$$ $$= e^{ (\tfrac{\sin ^2x}{1 - \sin ^2x})\ln3}$$ $$\displaystyle = e^{(\tan ^2x)\ln3}$$ $$= e^{\ln3^{\tan ^2x}}$$ $$= 3^{\tan ^2x}$$  Now roots of given quadratic equation is 1 and 27 and  also $$x\epsilon (0, \pi/2)$$, So $$\displaystyle 3^{\tan ^2x} = 27$$ $$\Rightarrow \tan x = \sqrt{3}$$ $$\Rightarrow \sin x =\dfrac{\sqrt{3}}{2}$$ and $$\cos x =\dfrac{1}{2}$$ So, $$\displaystyle ( \sin x + \cos x )^{-1} = \dfrac{2}{\sqrt{3} + 1}$$ $$= \dfrac{2}{\sqrt{3} + 1} .\dfrac{\sqrt{3} - 1}{\sqrt{3} - 1}$$ $$= \sqrt{3} -1$$ Hence, option 'A' is correct.Maths

Suggest Corrections

0

Similar questions
View More

People also searched for
View More