Question

Prove that $\frac{sin 2x}{(sec x + 1)} . \frac{sec 2x}{(sec 2x + 1)} = tan (\frac{x}{2})$

Answer 1

Explanation:

The given question involves trigonometric functions.

I have solved it using simple trigonometric conversions and manipulations.

I have also used some simple mathematical operations like addition, reciprocal and LCM etc.

I have given a detailed answer.

Kindly see the Attachment for the detailed answer.

I hope it will help you.

Answer 2

Question :- prove that sin 2x/sec( x+1) ×sec 2x/sec (2x+1) = tan(x/2) ?

Solution :-

solving LHS,

→ {sin2x/(secx+1)} * {sec2x/(sec2x+1)}

putting :-

• sec x = 1/cos x
• sec 2x = 1/cos 2x

→ {sin2x/(1/cosx)+1} * {(1/cos2x)/(1/cos2x)+1}

→ {2sinxcos²x/(1+cosx)} * {1/(1+cos2x)}

putting :-

• cos x = 2cos² x/2 - 1
• cos 2x = 2cos² x - 1

→ {2sinxcos²x/(1 + 2cos²x/2 - 1)} * {1/(1+2cos²x - 1)}

→ {2sinxcos²x/(2cos²x/2)} * {1/(2cos²x)}

→ sinx/(2cos²x/2)

putting :-

• sin x = 2 * sin x/2 * cos x /2

→ (2sinx/2cosx/2)/(2cos²x/2)

→ (sinx/2)/(cosx/2)

using :-

• sin x / cos x = tan x

→ (tanx/2) = RHS (Proved.)