Question

$\frac{ \sin(2a) }{1 + \cos(2a) } =$
(A)tanA
(B)cotA
(C)sinA
(D)cosA
​

Answer 1

★ Question :

$\bullet\ \; \sf \red{ \dfrac{sin(2A)}{1+cos(2A)} =\ ?}$

A. tan A  ✅

B. cot A

C. sin A

D. cos A

★ Solution :

We know that ,

$\bullet\ \; \sf \pink{sin(2 \theta )=2\ sin \theta .cos \theta}$

$\bullet\ \; \sf \green{1+cos(2 \theta)= 2cos^2 \theta}$

Sub. above formulas in our eq. , we get ,

$:\implies \sf \dfrac{sin(2A)}{1+cos(2A)}$

$:\implies \sf \dfrac{2\ .\ sin\ A\ .\ cos\ A}{2\ cos^2A}$

$:\implies \sf \dfrac{\cancel{2}\ sin\ A\ .\cancel{cos\ A}}{\cancel{2}\ cos^{\cancel{2}}A}$

$:\implies \sf \dfrac{sin\ A}{cos\ A}$

$:\implies \sf tan\ A\ \; \orange{\bigstar}$

Option (A)

Answer 2

Question :-

$\frac{sin(2a)}{1+cos(2a)}$

(A) $tanA$ ✔

(B) $cotA$

(C) $sinA$

(D) $cosA$

We know that :-

• $sin2\theta=2sin\theta cos\theta$
• $1+cos2\theta=2cos^{2}\theta$

Answer :-

$\frac{sin(2a)}{1+cos(2a)}$

$=>\frac{2sinA*cosA}{2cos^{2}A}$

$=>\frac{2*sinA*cosA}{2*cosA*cosA}$

$=>\frac{sinA}{cosA}$

$=>tanA$

Hence,

The value of $\frac{sin(2a)}{1+cos(2a)}$ is $tanA$.

☆ Hope it Helps ☆

$@NightRead$