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प्रश्न
Prove the following identities:
`secA/(secA + 1) + secA/(secA - 1) = 2cosec^2A`
Prove the following:
`secA/(secA + 1) + secA/(secA - 1) = 2"cosec"^2A`
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उत्तर
L.H.S. = `secA/(secA + 1) + secA/(secA - 1)`
= `(sec^2A - secA + sec^2A + secA)/(sec^2A - 1`
= `(2sec^2A)/tan^2A` ...(∵ sec2 A – 1 = tan2 A)
= `(2/cos^2A)/(sin^2A/cos^2A)`
= `2/sin^2A`
= 2 cosec2 A = R.H.S.
संबंधित प्रश्न
`(sec theta -1 )/( sec theta +1) = ( sin ^2 theta)/( (1+ cos theta )^2)`
Prove the following identities:
`sin theta/((cot theta + "cosec" theta)) - sin theta/((cot theta - "cosec" theta)) = 2`
What is the value of \[\frac{\tan^2 \theta - \sec^2 \theta}{\cot^2 \theta - {cosec}^2 \theta}\]
Prove the following identity :
`(sec^2θ - sin^2θ)/tan^2θ = cosec^2θ - cos^2θ`
Given `cos38^circ sec(90^circ - 2A) = 1` , Find the value of <A
Prove that `sqrt(2 + tan^2 θ + cot^2 θ) = tan θ + cot θ`.
Without using a trigonometric table, prove that
`(cos 70°)/(sin 20°) + (cos 59°)/(sin 31°) - 8sin^2 30° = 0`.
Prove that:
`(cos^3 θ + sin^3 θ)/(cos θ + sin θ) + (cos^3 θ - sin^3 θ)/(cos θ - sin θ) = 2`
tan2θ – sin2θ = tan2θ × sin2θ. For proof of this complete the activity given below.
Activity:
L.H.S. = `square`
= `square (1 - (sin^2θ)/(tan^2θ))`
= `tan^2θ (1 - square/((sin^2θ)/(cos^2θ)))`
= `tan^2θ (1 - (sin^2θ)/1 xx (cos^2θ)/square)`
= `tan^2θ (1 - square)`
= `tan^2θ xx square` ...[1 – cos2θ = sin2θ]
= R.H.S.
Prove that `(cot A - cos A)/(cot A + cos A) = (cos^2 A)/(1 + sin A)^2`
