Question

Prove that : (cosec p -sin p) (sec p-cos p) (tan p+ cot p ) = 1

Answer 1

Answer:

We have to prove, (cosec p -sin p) (sec p-cos p) (tan p+ cot p ) = 1

L.H.S.

$(cosec p -sin p) (sec p-cos p) (tan p+ cot p )$

= $(\frac{1}{sinp} - sinp)(\frac{1}{cosp}-cosp)(\frac{sinp}{cosp}+\frac{cosp}{sinp})$

= $(\frac{1-sin^2p}{sinp})(\frac{1-cos^2p}{cosp})(\frac{sin^2p+cos^2p}{sinpcosp})$

= $(\frac{cos^2p}{sinp})(\frac{sin^2p}{cosp})(\frac{1}{sinpcosp})$

= $\frac{cos^2p\times sin^2p}{sinp\times cosp\times sinp\times cos p}$

= $\frac{cos^2p\times sin^2p}{cos^2p\times sin^2p}$

= 1

= R.H.S.

Hence, proved.