Advertisements
Advertisements
प्रश्न
Prove the following identity :
`(secA - 1)/(secA + 1) = sin^2A/(1 + cosA)^2`
Advertisements
उत्तर
LHS = `(secA - 1)/(secA + 1)`
= `(1/cosA - 1)/(1/cosA + 1) = (1 - cosA)/(1 + cosA)`
= `(1 - cosA)/(1 + cosA) xx (1 + cosA)/(1 + cosA)`
= `(1-cos^2A)/(1 + cosA)^2`
= `sin^2A/(1 + cosA)^2` (∵ `1 - cos^2A = sin^2A`)
APPEARS IN
संबंधित प्रश्न
if `cos theta = 5/13` where `theta` is an acute angle. Find the value of `sin theta`
Prove the following identities:
`secA/(secA + 1) + secA/(secA - 1) = 2cosec^2A`
Prove the following identities:
`cosA/(1 - sinA) = sec A + tan A`
Prove the following identities:
`sqrt((1 - cosA)/(1 + cosA)) = sinA/(1 + cosA)`
If `(x/a sin a - y/b cos theta) = 1 and (x/a cos theta + y/b sin theta ) =1, " prove that "(x^2/a^2 + y^2/b^2 ) =2`
If` (sec theta + tan theta)= m and ( sec theta - tan theta ) = n ,` show that mn =1
Prove that:
tan (55° + x) = cot (35° – x)
Prove that `sqrt((1 + cos A)/(1 - cos A)) = (tan A + sin A)/(tan A. sin A)`
If x = h + a cos θ, y = k + b sin θ.
Prove that `((x - h)/a)^2 + ((y - k)/b)^2 = 1`.
Prove that cos2θ . (1 + tan2θ) = 1. Complete the activity given below.
Activity:
L.H.S = `square`
= `cos^2theta xx square .....[1 + tan^2theta = square]`
= `(cos theta xx square)^2`
= 12
= 1
= R.H.S
