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प्रश्न
Prove the following identities: cot θ - tan θ = `(2 cos^2 θ - 1)/(sin θ cos θ)`.
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उत्तर
LHS = cot θ - tan θ
= `cos θ/sin θ - sin θ/cos θ`
= `(cos^2 θ - sin^2 θ)/(sin θ. cos θ)`
= `(cos^2 θ - (1 - cos^2 θ))/(sin θ. cos θ)`
= `(2cos^2 θ - 1)/(sin θ. cos θ)`
= RHS
Hence proved.
संबंधित प्रश्न
Without using trigonometric tables evaluate
`(sin 35^@ cos 55^@ + cos 35^@ sin 55^@)/(cosec^2 10^@ - tan^2 80^@)`
Prove the following trigonometric identities.
`(cos theta - sin theta + 1)/(cos theta + sin theta - 1) = cosec theta + cot theta`
Prove the following identities:
(1 + cot A – cosec A)(1 + tan A + sec A) = 2
`cot theta/((cosec theta + 1) )+ ((cosec theta +1 ))/ cot theta = 2 sec theta `
If `(cosec theta - sin theta )= a^3 and (sec theta - cos theta ) = b^3 , " prove that " a^2 b^2 ( a^2+ b^2 ) =1`
Prove the following identity :
`1/(cosA + sinA - 1) + 2/(cosA + sinA + 1) = cosecA + secA`
Prove the following identity :
`(cosecθ)/(tanθ + cotθ) = cosθ`
If A + B = 90°, show that `(sin B + cos A)/sin A = 2tan B + tan A.`
Prove that `cot^2 "A" [(sec "A" - 1)/(1 + sin "A")] + sec^2 "A" [(sin"A" - 1)/(1 + sec"A")]` = 0
If sec θ = `25/7`, find the value of tan θ.
Solution:
1 + tan2 θ = sec2 θ
∴ 1 + tan2 θ = `(25/7)^square`
∴ tan2 θ = `625/49 - square`
= `(625 - 49)/49`
= `square/49`
∴ tan θ = `square/7` ........(by taking square roots)
