Advertisements
Advertisements
प्रश्न
Prove that:
(1 + tan A . tan B)2 + (tan A – tan B)2 = sec2 A sec2 B
Advertisements
उत्तर
L.H.S. = (1 + tanA . tan B)2 + (tan A – tan B)2
= 1 + tan2A . tan2B + 2 tan A . tan B + tan2A + tan2B – 2 tan A tan B
= 1 + tan2A + tan2B + tan2A + tan2B
= sec2A + tan2 B(1 + tan2A)
= sec2A + tan2 B sec2 A
= sec2A(1 + tan2B)
= sec2A sec2B
= R.H.S.
संबंधित प्रश्न
Prove the following identities:
`sqrt((1 - cosA)/(1 + cosA)) = cosec A - cot A`
Write the value of `(1 + cot^2 theta ) sin^2 theta`.
Prove the following identity :
secA(1 + sinA)(secA - tanA) = 1
Prove the following identity :
`cosA/(1 - tanA) + sinA/(1 - cotA) = sinA + cosA`
Prove the following identity :
`cosA/(1 - tanA) + sin^2A/(sinA - cosA) = cosA + sinA`
Prove the following identity :
`tan^2θ/(tan^2θ - 1) + (cosec^2θ)/(sec^2θ - cosec^2θ) = 1/(sin^2θ - cos^2θ)`
Prove that sec θ. cosec (90° - θ) - tan θ. cot( 90° - θ ) = 1.
If sinA + sin2A = 1, then the value of the expression (cos2A + cos4A) is ______.
If sin A = `1/2`, then the value of sec A is ______.
Factorize: sin3θ + cos3θ
Hence, prove the following identity:
`(sin^3θ + cos^3θ)/(sin θ + cos θ) + sin θ cos θ = 1`
