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प्रश्न
Prove that:
`1/(cosA + sinA - 1) + 1/(cosA + sinA + 1) = cosecA + secA`
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उत्तर
L.H.S. = `1/(cosA + sinA - 1) + 1/(cosA + sinA + 1)`
= `(cosA + sinA + 1 + cosA + sinA - 1)/((cosA + sinA)^2 - 1)`
= `(2(cosA + sinA))/(cos^2A + sin^2A + 2cosAsinA - 1)`
= `(2(cosA + sinA))/(1 + 2cosAsinA - 1)`
= `(cosA + sinA)/(cosAsinA)`
= `cosA/(cosAsinA) + sinA/(cosAsinA)`
= `1/sinA + 1/cosA`
= cosec A + sec A = R.H.S.
संबंधित प्रश्न
If secθ + tanθ = p, show that `(p^{2}-1)/(p^{2}+1)=\sin \theta`
If m=(acosθ + bsinθ) and n=(asinθ – bcosθ) prove that m2+n2=a2+b2
Prove the following trigonometric identities.
`(1 + sin θ)/cos θ+ cos θ/(1 + sin θ) = 2 sec θ`
If tan A = n tan B and sin A = m sin B , prove that `cos^2 A = ((m^2-1))/((n^2 - 1))`
Write the value of `sin theta cos ( 90° - theta )+ cos theta sin ( 90° - theta )`.
Write the value of ` sin^2 theta cos^2 theta (1+ tan^2 theta ) (1+ cot^2 theta).`
If `cot theta = 1/ sqrt(3) , "write the value of" ((1- cos^2 theta))/((2 -sin^2 theta))`
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 2(sin6A + cos6A) – 3(sin4A + cos4A) + 1 = 0
Prove the following:
`sintheta/(1 + cos theta) + (1 + cos theta)/sintheta` = 2cosecθ
