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प्रश्न
Prove that: `1/(sec θ - tan θ) = sec θ + tan θ`.
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उत्तर
LHS = `1/(sec θ - tan θ)`
= `1/((1/cos θ) - (sin θ/cos θ))`
= `(cos θ xx (1 + sin θ))/((1 - sin θ) xx ( 1 + sin θ))`
= `(cos θ( 1 + sin θ))/(1 - sin^2 θ)`
= `(cos θ( 1 + sin θ))/(cos^2 θ)`
= `1/cos θ + sin θ/cos θ`
= sec θ + tan θ
= RHS
Hence proved.
संबंधित प्रश्न
Prove the following trigonometric identities:
`(\text{i})\text{ }\frac{\sin \theta }{1-\cos \theta }=\text{cosec}\theta+\cot \theta `
Prove the following trigonometric identities.
`"cosec" theta sqrt(1 - cos^2 theta) = 1`
Prove the following identities:
`(1+ sin A)/(cosec A - cot A) - (1 - sin A)/(cosec A + cot A) = 2(1 + cot A)`
` tan^2 theta - 1/( cos^2 theta )=-1`
Write the value of `(cot^2 theta - 1/(sin^2 theta))`.
Prove the following identity :
`(sinA - sinB)/(cosA + cosB) + (cosA - cosB)/(sinA + sinB) = 0`
Prove that ( 1 + tan A)2 + (1 - tan A)2 = 2 sec2A
Prove the following identities:
`(1 - tan^2 θ)/(cot^2 θ - 1) = tan^2 θ`.
If tan θ – sin2θ = cos2θ, then show that sin2 θ = `1/2`.
Complete the following activity to prove:
cotθ + tanθ = cosecθ × secθ
Activity: L.H.S. = cotθ + tanθ
= `cosθ/sinθ + square/cosθ`
= `(square + sin^2theta)/(sinθ xx cosθ)`
= `1/(sinθ xx cosθ)` ....... ∵ `square`
= `1/sinθ xx 1/cosθ`
= `square xx secθ`
∴ L.H.S. = R.H.S.
