The value of $(2 cos^2 θ − 1) [\frac{1+tan θ}{1-tan θ}+\frac{1-tan θ}{1+tan θ}]$ is:
Answer & explanation
Correct answer: option 1
( 2 cos²θ - 1 ) . [ \(\frac{ 1 + tanθ }{1 - tanθ}\) + \(\frac{ 1 - tanθ }{1 + tanθ}\) ]
{ we know, 2 cos²θ - 1 = cos 2θ }
= ( cos2θ) × \(\frac{ (1 + tanθ)² + (1+ tanθ)² }{(1 - tanθ).(1 + tanθ) }\)
= ( cos2θ) × \(\frac{ 1 + tan² θ+ 2 tanθ + 1 + tan² θ- 2 tanθ }{ 1 - tan² θ }\)
= ( cos2θ) × \(\frac{ 1 + tan² θ + 1 + tan² θ }{ 1 - tan² θ }\)
= ( cos2θ) × \(\frac{ 2( 1 + tan² θ ) }{ 1 - tan² θ }\)
= 2 ( cos2θ) × \(\frac{ ( cos² θ + sin² θ ) }{ cos² θ - sin² θ }\)
{ we know, cos² θ + sin² θ = 1 }
= 2( cos2θ) × \(\frac{ 1 }{ cos² θ - sin² θ }\)
= 2( cos2θ) × \(\frac{ 1 }{ cos2θ }\)
= 2