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प्रश्न
Prove the following identities:
`( i)sin^{2}A/cos^{2}A+\cos^{2}A/sin^{2}A=\frac{1}{sin^{2}Acos^{2}A)-2`
`(ii)\frac{cosA}{1tanA}+\sin^{2}A/(sinAcosA)=\sin A\text{}+\cos A`
`( iii)((1+sin\theta )^{2}+(1sin\theta)^{2})/cos^{2}\theta =2( \frac{1+sin^{2}\theta}{1-sin^{2}\theta } )`
उत्तर
(i)We have,
`LHS=sin^{2}A/cos^{2}A+cos^{2}A/sin^{2}A =\frac{sin^{4}A+cos^2A}{sin^{2}Acos^{2}A}`
`=((\sin ^{2}A)^{2}+(\cos ^{2}A)^{2}+2\sin^{2}A\cos ^{2}A-2\sin^{2}Acos^{2}A)/(\sin^{2}A\cos ^{2}A)`
`=((\sin ^{2}A+\cos ^{2}A)^{2}-2\sin ^{2}A\cos^{2}A)/(sin ^{2}A\cos ^{2}A`
`=(1-2\sin ^{2}A\cos ^{2}A)/(\sin^{2}A\cos ^{2}A)`
`=1/(\sin^{2}A\cos ^{2}A)-2=RHS`
(ii) We have,
`LHS=\frac{\cos A}{1-\tan A}+\sin ^{2}A/(\sin A-\cos A)`
`=\frac{\cos A}{1-\frac{\sin A}{\cos A}}+\sin^{2}A/(\sinA-\cos A)`
`=\frac{\cos A}{\frac{\cos A-\sin A}{\cos A}}+\sin ^{2}A/(\sinA-\cos A)`
`=\cos ^{2}A/(\cos A\sin A)+sin ^{2}A/(\sin A\cos A)`
`=\cos ^{2}A/(\cos A\sin A)-\sin ^{2}A/(\cos A\sin A)`
`=(\cos ^{2}A-\sin ^{2}A)/{\cos A-\sin A}`
`=\frac{(\cos A+\sin A)\,(\cos A-\sin A)}{\cos A-\sin A}`
= cos A + sin A = RHS
(iii) We have,
`LHS=((1+\sin \theta )^{2}+(1\sin \theta )^{2})/(\cos^{2}\theta )`
`=\frac{(1+2\sin \theta +\sin ^{2}\theta )+(12\sin \theta +\sin^{2}\theta )}{\cos ^{2}theta `
`=(2+2\sin^{2}\theta)/ \cos ^{2}\theta=(2(1+\sin^{2}\theta ))/{1-\sin ^{2}\theta }=2( \frac{1+\sin^{2}\theta }{1-\sin ^{2}\theta } )`
= RHS.