Advertisements
Advertisements
प्रश्न
Prove that ( 1 + tan A)2 + (1 - tan A)2 = 2 sec2A
Advertisements
उत्तर
LHS = ( 1 + tan A)2 + (1 - tan A)2
= 1 + 2 tan A + tan2A + 1 - 2 tan A + tan2A
= 2( 1 + tan2A)
= 2 sec2A
= RHS
Hence proved.
संबंधित प्रश्न
Prove the following trigonometric identities.
`1/(sec A - 1) + 1/(sec A + 1) = 2 cosec A cot A`
Prove the following trigonometric identities.
(sec A − cosec A) (1 + tan A + cot A) = tan A sec A − cot A cosec A
Prove that:
`(tanA + 1/cosA)^2 + (tanA - 1/cosA)^2 = 2((1 + sin^2A)/(1 - sin^2A))`
Prove that:
`cosA/(1 + sinA) = secA - tanA`
If \[\sin \theta = \frac{1}{3}\] then find the value of 9tan2 θ + 9.
If 5x = sec θ and \[\frac{5}{x} = \tan \theta\]find the value of \[5\left( x^2 - \frac{1}{x^2} \right)\]
Prove the following identity :
`1/(sinA + cosA) + 1/(sinA - cosA) = (2sinA)/(1 - 2cos^2A)`
Prove that `((1 + sin θ - cos θ)/( 1 + sin θ + cos θ))^2 = (1 - cos θ)/(1 + cos θ)`.
If tan θ + cot θ = 2, then tan2θ + cot2θ = ?
Prove that `(sec A)/(tan A + cot A) = sin A`.
