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प्रश्न
`cos^2 theta /((1 tan theta))+ sin ^3 theta/((sin theta - cos theta))=(1+sin theta cos theta)`
उत्तर
`cos^2 theta /((1 tan theta))+ sin ^3 theta/((sin theta - cos theta))=(1+sin theta cos theta)`
LHS=`cos^2theta/((1-tan theta))+sin ^3theta/((sin theta - cos theta))`
=`cos^2theta/(1-sintheta/costheta)+sin^3 theta/((sin theta-costheta))`
=`cos^3 theta/((cos theta-sin theta))+ sin ^3 theta/((sintheta-cos theta))`
=`(cos^3theta-sin^3 theta)/((costheta - sin theta))`
=`((cos theta-sintheta)(cos^2 theta+cos theta sin +sin^2theta))/((costheta-sintheta))`
=`(sin^2theta + cos^2 theta + cos theta sin theta)`
=`(1+sin theta cos theta)`
=RHS
Hence, L.H.S = R.H.S.
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संबंधित प्रश्न
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(1 + cos α) (1 + cos β) (1 + cos γ) = (1 − cos α) (1 − cos α) (1 − cos β) (1 − cos γ)
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(sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A
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`(1 + sinA)/(1 - sinA) = (cosecA + 1)/(cosecA - 1)`
Prove the following identity :
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Prove that: `(sin A + cos A)/(sin A - cos A) + (sin A - cos A)/(sin A + cos A) = 2/(sin^2 A - cos^2 A)`.
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