Prove that :
( tan A + cot A ) sin A×cos A = 1
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Given: LHS = ( tan A + cot A ) sin A×cos A
RHS = 1
Taking LHS,
= ( tan A + cot A ) sin A×cos A
= ( sin A / cos A + cos A / sin A ) sin A×cos A
[∵ tan A = sin A/ cos A and cot A = cos A/ sin A]
= {(sin²A + cos²A)/ sin A × cos A} sin A×cos A [∵ by taking LCM]
= (1 / sin A×cos A ) sin A×cos A [∵sin²A + cos²A = 1]
= 1
∴ So here LHS = RHS i.e, ( tan A + cot A ) sin A×cos A = 1
hence proved..
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