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Find the value of sin thetha×Cos3thetha×cosec2thetha-2Cos square thetha


2 years ago

Susmita
425 Points
							Hey I am using A in place of theta.Given sinAcos3Acosec2A-2cos2A$=\frac{sinAcos(A+2A)}{sin2A} -2cos^2A$$=\frac{sinA(cosAcos2A-sinAsin2A)}{sin2A} -2cos^2A$$=\frac{sinAcosAcos2A}{sin2A} -\frac{sinAsin2AsinA}{sin2A}-2cos^2A$$=\frac{sinAcosAcos2A}{2sinAcosA} -sin^2A-2cos^2A$ $=\frac{1+2cos^2A}{2} -sin^2A-2cos^2A$$=\frac{1}{2}+cos^2A -sin^2A-2cos^2A$$=\frac{1}{2}-cos^2A -sin^2A$$=\frac{1}{2}-1=-\frac{1}{2}$I have used only basic formulas to obtain this.If you find any hardle ask me.Hope this helps

2 years ago
Yash
20 Points
							Let Theta=¶ Sin¶XCos3¶XCosec2¶-Cos²¶=Sin¶X(4Cos³¶-3Cos¶)X1/sin2¶-Cos²¶=Sin¶X(4Cos³¶-3Cos¶)X1/2sin¶cos¶-Cos²¶=2Cos²¶-3/2-2Cos²¶=-3/2

2 years ago
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