Advertisements
Advertisements
Question
Prove the following identity :
`(1 + tan^2A) + (1 + 1/tan^2A) = 1/(sin^2A - sin^4A)`
Advertisements
Solution 1
...
Solution 2
LHS = `(1 + tan^2A) + (1 + 1/tan^2A)`
= `(1 + sin^2A/cos^2A) + (1 + 1/(sin^2A/cos^2A))`
= `((cos^2A + sin^2A)/(cos^2A)) + ((cos^2A + sin^2A)/(sin^2A))`
= `1/(1 - sin^2A) + 1/sin^2A` (∵ `cos^2A + sin^2A = 1`)
= `(sin^2A + 1 - sin^2A)/(sin^2A(1 - sin^2A)) = 1/(sin^2A - sin^4A)`
APPEARS IN
RELATED QUESTIONS
If `sec alpha=2/sqrt3` , then find the value of `(1-cosecalpha)/(1+cosecalpha)` where α is in IV quadrant.
(i)` (1-cos^2 theta )cosec^2theta = 1`
If secθ + tanθ = m , secθ - tanθ = n , prove that mn = 1
Choose the correct alternative:
1 + tan2 θ = ?
Find the value of ( sin2 33° + sin2 57°).
Prove that `(sec θ - 1)/(sec θ + 1) = ((sin θ)/(1 + cos θ ))^2`
Prove that `cot^2 "A" [(sec "A" - 1)/(1 + sin "A")] + sec^2 "A" [(sin"A" - 1)/(1 + sec"A")]` = 0
sec2θ – tan2θ = ?
Prove that `(cot A)/(1 - tan A) + (tan A)/(1 - cot A) = 1 + tan A + cot A = sec A . "cosec" A + 1`.
If sin A = `1/2`, then the value of sec A is ______.
