Question

IF $f\left(\theta \right)=\frac{1-\sin 2\theta +\cos 2\theta }{2\cos 2\theta }$ then the value of $f\left({11}^{\circ }\right).$$f\left({34}^{\circ }\right)$

• A. $\frac{1}{2}$
• B. $\frac{3}{4}$
• C. $\frac{1}{4}$
• D. $1$

Answer 1

Answer: (B)

$\frac{3}{4}$

$f\left(\theta \right)=\frac{1-sin2\theta +cos2\theta }{2cos2\theta }$

$=\frac{(1+cos2\theta )-sin2\theta }{2cos2\theta }$

$=\frac{2co{s}^2\theta -2sin\theta cos\theta }{2(co{s}^2\theta -si{n}^2\theta )}$

$=\frac{2cos\theta (cos\theta -sin\theta )}{2(cos\theta -sin\theta )(cos\theta +sin\theta )}$

$=\frac{cos\theta }{cos\theta +sin\theta }=\frac{1}{1+tan\theta }$

now $f\left({11}^{\circ }\right)\times f\left({34}^{\circ }\right)=\frac{1}{1+tan{11}^{\circ }}\times \frac{1}{1+tan{34}^{\circ }}$

$=\frac{1}{1+tan{11}^{\circ }}\times \frac{1}{1+tan({45}^{\circ }-{11}^{\circ })}$

$=\frac{1}{1+tan{11}^{\circ }}\times \frac{1}{1+\frac{tan{45}^{\circ }-tan{11}^{\circ }}{1+tan{45}^{\circ }tan{11}^{\circ }}}$

$=\frac{1}{1+tan{11}^{\circ }}\times \frac{1}{1+\frac{1-tan{11}^{\circ }}{1+tna{11}^{\circ }}}=\frac{1}{2}$