solve this one
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LHS = tan A /sec²A + cotA /cosec²A
= (sinA/cosA) /(1/cos²A) + (cosA /sinA)/(1/sin²A)
= sinAcosA + cosA sinA
hence ,answer will be 2 sinAcosA
= (sinA/cosA) /(1/cos²A) + (cosA /sinA)/(1/sin²A)
= sinAcosA + cosA sinA
hence ,answer will be 2 sinAcosA
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