Question #d1404

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Answer 1 · 2017-11-22T12:40:22

Please refer to Explanation.

We have, #sec^2theta+csc^2theta=1/cos^2theta+1/sin^2theta,#

#=(sin^2theta+cos^2theta)/(cos^2thetasin^2theta),#

#1/(cos^2thetasin^2theta),#

#=1/cos^2theta*1/sin^2theta,#

#sec^2thetacsc^2theta.#

Answer 2 · 2017-11-22T12:40:48

See explanation.

#LHS=sec^2x+csc^2x=1/cos^2x+1/sin^2x=(sin^2x+cos^2x)/(sin^2x*cos^2x)#

#=1/(sin^2xcos^2x)=1/sin^2x*1/cos^2x=csc^2x*sec^2x=RHS#