Advertisements
Advertisements
प्रश्न
Prove the following identity :
`(1 - tanA)^2 + (1 + tanA)^2 = 2sec^2A`
Advertisements
उत्तर
LHS = `(1 - tanA)^2 + (1 + tanA)^2`
= `1 + tan^2A - 2tanA + 1 + tan^2A + 2tanA`
= `2(1 + tan^2A) = 2sec^2A` = RHS
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities.
`(cos theta - sin theta + 1)/(cos theta + sin theta - 1) = cosec theta + cot theta`
Prove that:
cos A (1 + cot A) + sin A (1 + tan A) = sec A + cosec A
`sec theta (1- sin theta )( sec theta + tan theta )=1`
`(1+ cos theta - sin^2 theta )/(sin theta (1+ cos theta))= cot theta`
`(1+ tan theta + cot theta )(sintheta - cos theta) = ((sec theta)/ (cosec^2 theta)-( cosec theta)/(sec^2 theta))`
If `sec theta + tan theta = x," find the value of " sec theta`
Prove that `(sinθ - cosθ + 1)/(sinθ + cosθ - 1) = 1/(secθ - tanθ)`
Prove the following identity :
`cos^4A - sin^4A = 2cos^2A - 1`
Prove that `sqrt((1 + sin A)/(1 - sin A))` = sec A + tan A.
Prove that `(cot A)/(1 - tan A) + (tan A)/(1 - cot A) = 1 + tan A + cot A = sec A . "cosec" A + 1`.
