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

5k Whatsapp video , 6 Years ago
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Susmita

Last Activity: 6 Years ago

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

Yash

Last Activity: 6 Years ago

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

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