prove that sin (180+A)*cos(90-A)*tan(270-A)/sec(540-A)*cos(360+A)*cosec(270+A)=-sin A cos^2 A
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To solve this sum, follow this method
LHS>>>
-sinA× secA×cotA/secA× sinA ×cosA
-sinA× secA/cosA
=sinA×cosA×cosA
=-sinA00×cos^2A
LHS>>>
-sinA× secA×cotA/secA× sinA ×cosA
-sinA× secA/cosA
=sinA×cosA×cosA
=-sinA00×cos^2A
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