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प्रश्न
Prove the following identity :
`(1 + cotA + tanA)(sinA - cosA) = secA/(cosec^2A) - (cosecA)/sec^2A`
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उत्तर
LHS = `(1 + cotA + tanA)(sinA - cosA)`
= `(1 + cosA/sinA + sinA/cosA)(sinA - cosA)`
= `((sinAcosA + cos^2A + sin^2A)/(sinAcosA))(sinA - cosA)`
= `((sin^3A - cos^3A))/(sinAcosA)` (∵(`sin^3A - cos^3A) = (sinA - cosA)(sinA cosA + cos^2A + sin^2A`))
= `sin^3A/(sinAcosA) - cos^3A/(sinAcosA)`
= `sin^2A/cosA - cos^2A/sinA = 1/cosA xx sin^2A - 1/sinA xx cos^2A`
= `secAsin^2A - cosecAcos^2A`
= `secA/(cosec^2A) - (cosecA)/sec^2A`
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संबंधित 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 } )`
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`(1+ tan^2 theta)/(1+ tan^2 theta)= (cos^2 theta - sin^2 theta)`
The value of \[\sqrt{\frac{1 + \cos \theta}{1 - \cos \theta}}\]
Prove that tan2Φ + cot2Φ + 2 = sec2Φ.cosec2Φ.
Prove that sin θ sin( 90° - θ) - cos θ cos( 90° - θ) = 0
Prove that sin (90° - θ) cos (90° - θ) = tan θ. cos2θ.
If cos θ = `24/25`, then sin θ = ?
Prove that `1/("cosec" theta - cot theta)` = cosec θ + cot θ
tan2θ – sin2θ = tan2θ × sin2θ. For proof of this complete the activity given below.
Activity:
L.H.S = `square`
= `square (1 - (sin^2theta)/(tan^2theta))`
= `tan^2theta (1 - square/((sin^2theta)/(cos^2theta)))`
= `tan^2theta (1 - (sin^2theta)/1 xx (cos^2theta)/square)`
= `tan^2theta (1 - square)`
= `tan^2theta xx square` .....[1 – cos2θ = sin2θ]
= R.H.S
