Advertisements
Advertisements
प्रश्न
Prove that cosec2 (90° - θ) + cot2 (90° - θ) = 1 + 2 tan2 θ.
उत्तर
LHS = cosec2 (90° - θ) + cot2 (90° - θ)
= sec2 θ + tan2θ
= 1 + tan2θ + tan2θ
= 1 + 2 tan2θ
= RHS
Hence proved.
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities.
`(cot^2 A(sec A - 1))/(1 + sin A) = sec^2 A ((1 - sin A)/(1 + sec A))`
Prove the following identities:
cosec A(1 + cos A) (cosec A – cot A) = 1
Prove the following identities:
`sinA/(1 + cosA) = cosec A - cot A`
Prove the following identities:
`cosA/(1 - sinA) = sec A + tan A`
Prove that `(sinθ - cosθ + 1)/(sinθ + cosθ - 1) = 1/(secθ - tanθ)`
What is the value of \[6 \tan^2 \theta - \frac{6}{\cos^2 \theta}\]
Write True' or False' and justify your answer the following :
The value of \[\cos^2 23 - \sin^2 67\] is positive .
If cosθ + sinθ = `sqrt2` cosθ, show that cosθ - sinθ = `sqrt2` sinθ.
If 4 tanβ = 3, then `(4sinbeta-3cosbeta)/(4sinbeta+3cosbeta)=` ______.
If tan α + cot α = 2, then tan20α + cot20α = ______.