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प्रश्न
`(sin theta +cos theta )/(sin theta - cos theta)+(sin theta- cos theta)/(sin theta + cos theta) = 2/((sin^2 theta - cos ^2 theta)) = 2/((2 sin^2 theta -1))`
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उत्तर
We have , `(sin theta +cos theta )/(sin theta - cos theta)+(sin theta- cos theta)/(sin theta + cos theta) `
=`((sin theta + cos theta )^2 + (sin theta - cos theta)^2) /((sin theta - cos theta )(sin theta + cos theta))`
=`(sin^2 theta + cos ^2 theta + 2 sin theta cos theta + sin^2 theta + cos^2 theta -2 sin theta cos theta)/(sin^2 theta - cos ^2 theta)`
=`(1+1)/(sin^2 theta - cos^2 theta)`
=`2/(sin^2 theta - cos^2 theta)`
Again ,` 2/(sin^2 theta - cos^2 theta)`
=`2/(sin^2 theta -(1-sin^2 theta))`
=`2/(2 sin ^2 theta -1)`
APPEARS IN
संबंधित प्रश्न
If tanθ + sinθ = m and tanθ – sinθ = n, show that `m^2 – n^2 = 4\sqrt{mn}.`
Prove the following identities:
`cosA/(1 - sinA) = sec A + tan A`
Prove the following identities:
`1 - cos^2A/(1 + sinA) = sinA`
Prove the following identities:
`cot^2A((secA - 1)/(1 + sinA)) + sec^2A((sinA - 1)/(1 + secA)) = 0`
If 4 cos2 A – 3 = 0, show that: cos 3 A = 4 cos3 A – 3 cos A
`cot^2 theta - 1/(sin^2 theta ) = -1`a
If `(cot theta ) = m and ( sec theta - cos theta) = n " prove that " (m^2 n)(2/3) - (mn^2)(2/3)=1`
If x = a sin θ and y = bcos θ , write the value of`(b^2 x^2 + a^2 y^2)`
If `sin theta = x , " write the value of cot "theta .`
Write True' or False' and justify your answer the following :
The value of \[\sin \theta\] is \[x + \frac{1}{x}\] where 'x' is a positive real number .
Prove the following identity :
`(1 - cos^2θ)sec^2θ = tan^2θ`
Prove the following identity :
`((1 + tan^2A)cotA)/(cosec^2A) = tanA`
Find the value of `θ(0^circ < θ < 90^circ)` if :
`tan35^circ cot(90^circ - θ) = 1`
Prove that:
`(sin A + cos A)/(sin A - cos A) + (sin A - cos A)/(sin A + cos A) = 2/(2 sin^2 A - 1)`
If sin θ + cos θ = a and sec θ + cosec θ = b , then the value of b(a2 – 1) is equal to
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)
Prove that `(cosθ)/(1 + sinθ) = (1 - sinθ)/(cosθ)`.
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.
