Question

Tan(π\4+x)+tan(π\4-x)=2sec2x

Answer 1

Step-by-step explanation:

Answer:

Step-by-step explanation:

The given equation is:

tan({\frac{\pi}{4}+x)+tan({\frac{\pi}{4}-x)=2sec2x

Taking the LHS of the above equation, we get

tan({\frac{\pi}{4}+x)+tan({\frac{\pi}{4}-x)=\frac{1+tanx}{1-tanx}+\frac{1-tanx}{1+tanx}

=\frac{(1+tanx)^{2}+(1-tanx)^2}{1-tan^2x}

1−tan

2

x

(1+tanx)

2

+(1−tanx)

2

=\frac{2(1+tan^2x)}{(1-tan^2x)}

(1−tan

2

x)

2(1+tan

2

x)

=\frac{2(cos^2x+sin^2x)}{cos^2x-sin^2x}

cos

2

x−sin

2

x

2(cos

2

x+sin

2

x)

=\frac{2}{cos2x}

cos2x

2

=2sec2x2sec2x

=RHS

Hence proved