Question

निम्नलिखित को सिद्ध कीजिए: $\dfrac {\sin x + \sin 3x}{ \cos x + \cos 3x} = \tan 2x$

Answer 1

Prove that : $\dfrac {\sin x + \sin 3x}{ \cos x + \cos 3x} = \tan 2x$



Sin x + Sin 3x ÷ Cos x + Cos 3x = tan 2x

2 Sin ( x + 3x ) /2 . Cos ( 3x - x ) /2 ÷ 2 Cos ( x + 3x )/2 . Cos ( 3x - x ) /2


2 Sin 2x . Cos x ÷ 2 Cos 2x. Cos x

Sin 2x . Cos x ÷ Cos 2x. Cos x

Sin 2x ÷ Cos 2x

tan 2x


Trigonometric Function Formula Used !


Sin A + Sin B = 2 Sin ( A + B)/2 COS (A - B)/2

Cos A + Cos B = 2 COS ( A+ B)/2 . Cos ( A - B )/2

Answer 2

= Sin x + Sin 3x ÷ Cos x + Cos 3x = tan 2x

= 2 Sin ( x + 3x ) /2 × Cos ( 3x - x ) /2 ÷ 2 Cos ( x + 3x )/2 × Cos ( 3x - x ) /2


= 2 Sin 2x . Cos x / 2 Cos 2x. Cos x

= Sin 2x . Cos x /Cos 2x. Cos x

= Sin 2x / Cos 2x

= tan 2x



Answer in Latex Version for better Results!

$Sin x + Sin 3x ÷ Cos x + Cos 3x = tan 2x \\ \\ = 2 Sin ( x + 3x ) /2 × Cos ( 3x - x ) /2 ÷ 2 Cos ( x + 3x )/2 × Cos ( 3x - x ) /2 \\ \\ = 2 Sin 2x . Cos x / 2 Cos 2x. Cos x \\ \\ = Sin 2x . Cos x /Cos 2x. Cos x \\ \\ = Sin 2x / Cos 2x \\ \\ = tan 2x$