JEE MATHEMATICS MCQ: TRIGONOMETRIC EQUATIONS

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Analyzing +4 | -1

\exp ^{\left[\left(\sin ^2 x+\sin ^4 x+\sin ^6 x+\ldots\right) \ln 2\right]}

satisfies the equation

y^2-9 y+8=0

, then the value of

\frac{\cos x}{\cos x+\sin x}, 0 < x < \frac{\pi}{2}

, is

View answer and explanation

Correct option: B

\frac{\sqrt{3}-1}{2}

Option B,

\frac{{\sqrt{{3}}-1}}{{2}}

, is correct because it is the simplified value of

\frac{{\cos x}}{{\cos x+\sin x}}

when

x = \frac{{\pi}}{{3}}

, which is obtained from solving the given equation.

More Questions from TRIGONOMETRIC EQUATIONS

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\sin \,\,x\,\,\;cos\,\,3x = \sin \,\,3x\,\,\cos \,\,5x

\left[ {0,\;\,\frac{\pi }{2}} \right]

\left( {x,\,y} \right)

x,\,\,y \in \left[ {0,10} \right]

\left( {\sqrt {{{\sin }^2} - \sin \,\,x + \frac{1}{2}} } \right){2^{{{\sec }^2}y}}\,\, \cdot \,\, \le 1

\mathrm{X}

\theta \rightarrow \frac{\pi}{2}

\sin x+\cos x=\frac{\sqrt{7}}{2}

x \in\left[0, \frac{\pi}{4}\right]

\tan \frac{x}{2}

\tan \frac{\pi}{5}+2 \tan \frac{2 \pi}{5}+4 \cot \frac{4 \pi}{5}

0 < x < \pi

\cos x+\sin x=\frac{1}{2}

\tan x