prove that (cosec a - sin a ) ( sec a - cos a ) = 1/tan a + cot a
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How can I prove that (cosec A-sin A) (sec A- cos A) =1/tan A+ cot A?
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We have to just simplify both side of the question.
Now, taking L. H. S =
(cosecA - sinA)×(secA - cosA)
= (1/sinA - sinA)×(1/cosA - cosA)
= (1-sin²A)/sinA × (1-cos²A)/cosA
= cos²A/sinA × sin²A/cosA
= cos²Asin²A/cosAsinA
= cosAsinA
Now, R. H. S =
1/(tanA + cotA)
= 1/(sinA/cosA + cosA/sinA)
= 1/[(sin²A + cos²A)/cosAsinA]
= 1/[(1/cosAsinA)]
= cosAsinA
Since L. H. S. = R. H. S.
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