Question

prove that tan(45+theta)+tan(45-theta) = 2sec2theta

Answer 1

tan(45°+θ)+tan(45°-θ)
=(tan45°+tanθ)/(1-tan45°tanθ)+(tan45°-tanθ)/(1+tan45°tanθ)
=(1+tanθ)/(1-tanθ)+(1-tanθ)/(1+tanθ)   [∵, tan45°=1]
={(1+tanθ)²+(1-tanθ)²}/(1-tanθ)(1+tanθ)
=(1+2tanθ+tan²θ+1-2tanθ+tan²θ)/(1-tan²θ)
=(2+2tan²θ)/(1-tan²θ)
=2(1+sin²θ/cos²θ)/(1-sin²θ/cos²θ)
=2{(cos²θ+sin²θ)/cos²θ}/{(cos²θ-sin²θ)/cos²θ}
=2(1/cos2θ)   [∵, sin²θ+cos²θ=1 and cos²θ-sin²θ=cos2θ]
=2sec2θ (Proved)

Answer 2

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