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प्रश्न
`sin theta (1+ tan theta) + cos theta (1+ cot theta) = ( sectheta+ cosec theta)`
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उत्तर १
LHS = ` sin theta (1+ tan theta ) + cos theta ( 1+ cot theta )`
=` sin theta + sin theta xx (sin theta)/(cos theta) + cos theta +cos theta xx (cos theta)/( sin theta)`
= `( cos theta sin ^2 theta + sin^3 theta + cos^2 theta sin theta + cos^3 theta)/(cos theta sin theta)`
=`((sin^3 theta + cos^3 theta)+(cos theta sin ^2 theta + cos ^2 theta sin theta))/(cos theta sin theta)`
=`((sin theta + cos theta )(sin^2 theta - sin theta cos theta + cos ^2 theta )+ sin theta cos theta ( sin theta + cos theta))/(cos theta sin theta)`
=`((sin theta + cos theta )( sin^2 theta + cos^2 theta - sin theta cos theta + sin theta cos theta))/(cos theta sin theta)`
=`((sin theta + cos theta)(1))/(cos theta sin theta)`
= `(sin theta)/(cos theta sin theta) + (cos theta)/( cos theta sin theta)`
=`1/cos theta + 1/ sin theta`
=` sec theta + cosec theta`
=RHS
उत्तर २
LHS = ` sin theta (1+ tan theta ) + cos theta ( 1+ cot theta )`
=` sin theta + sin theta xx (sin theta)/(cos theta) + cos theta +cos theta xx (cos theta)/( sin theta)`
= `( cos theta sin ^2 theta + sin^3 theta + cos^2 theta sin theta + cos^3 theta)/(cos theta sin theta)`
=`((sin^3 theta + cos^3 theta)+(cos theta sin ^2 theta + cos ^2 theta sin theta))/(cos theta sin theta)`
=`((sin theta + cos theta )(sin^2 theta - sin theta cos theta + cos ^2 theta )+ sin theta cos theta ( sin theta + cos theta))/(cos theta sin theta)`
=`((sin theta + cos theta )( sin^2 theta + cos^2 theta - sin theta cos theta + sin theta cos theta))/(cos theta sin theta)`
=`((sin theta + cos theta)(1))/(cos theta sin theta)`
= `(sin theta)/(cos theta sin theta) + (cos theta)/( cos theta sin theta)`
=`1/cos theta + 1/ sin theta`
=` sec theta + cosec theta`
=RHS
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We have, 1 + cot2θ = cosec2θ
1 + `square` = cosec2θ
1 + `square` = cosec2θ
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