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प्रश्न
\[\frac{\tan \theta}{\sec \theta - 1} + \frac{\tan \theta}{\sec \theta + 1}\] is equal to
विकल्प
2 tan θ
2 sec θ
2 cosec θ
2 tan θ sec θ
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उत्तर
The given expression is `tan θ /(secθ-1)+tan θ/(sec θ+1)`
=` (tan θ (sec θ+1)+tan θ(secθ-1))/((secθ-1)(secθ+1))`
= `(tan θ sec θ+tanθ+tan θ secθ-tan θ)/(sec^2θ-1)`
=`( 2tanθ secθ)/tan^2θ`
=`(2secθ)/tan θ`
= `(2 1/cos θ)/(sinθ/cos θ)`
=`2 1/ sinθ`
= `2 cosec θ`
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संबंधित प्रश्न
Prove the following trigonometric identities.
(1 + tan2θ) (1 − sinθ) (1 + sinθ) = 1
Prove the following trigonometric identities
`((1 + sin theta)^2 + (1 + sin theta)^2)/(2cos^2 theta) = (1 + sin^2 theta)/(1 - sin^2 theta)`
Prove the following trigonometric identities.
`(cos A cosec A - sin A sec A)/(cos A + sin A) = cosec A - sec A`
Prove the following trigonometric identities.
if x = a cos^3 theta, y = b sin^3 theta` " prove that " `(x/a)^(2/3) + (y/b)^(2/3) = 1`
If `(x/a sin a - y/b cos theta) = 1 and (x/a cos theta + y/b sin theta ) =1, " prove that "(x^2/a^2 + y^2/b^2 ) =2`
If` (sec theta + tan theta)= m and ( sec theta - tan theta ) = n ,` show that mn =1
Write the value of ` cosec^2 (90°- theta ) - tan^2 theta`
Find the value of sin ` 48° sec 42° + cos 48° cosec 42°`
Prove that:
`(sin^2θ)/(cosθ) + cosθ = secθ`
Prove that `(sinθ - cosθ + 1)/(sinθ + cosθ - 1) = 1/(secθ - tanθ)`
If sec θ + tan θ = x, write the value of sec θ − tan θ in terms of x.
Write the value of \[\cot^2 \theta - \frac{1}{\sin^2 \theta}\]
Prove the following identity:
`cosA/(1 + sinA) = secA - tanA`
Prove the following identity:
tan2A − sin2A = tan2A · sin2A
Prove the following identity :
`sqrt((1 - cosA)/(1 + cosA)) = sinA/(1 + cosA)`
Prove the following identity :
`1/(cosA + sinA - 1) + 2/(cosA + sinA + 1) = cosecA + secA`
Prove the following identity :
`(sec^2θ - sin^2θ)/tan^2θ = cosec^2θ - cos^2θ`
Prove that sin2A . tan A + cos2A . cot A + 2 sin A . cos A = tan A + cot A.
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.
Show that: `tan "A"/(1 + tan^2 "A")^2 + cot "A"/(1 + cot^2 "A")^2 = sin"A" xx cos"A"`
