Advertisements
Advertisements
Question
Prove that sec2 (90° - θ) + tan2 (90° - θ) = 1 + 2 cot2 θ.
Advertisements
Solution
LHS = sec2 (90° - θ) + tan2 (90° - θ)
= cosec2θ + cot2θ
= 1 + cot2θ + cot2θ
= 1 + 2cot2θ
= RHS
Hence proved.
RELATED QUESTIONS
Prove the following identities:
`(1 + sinA)/cosA + cosA/(1 + sinA) = 2secA`
If 4 cos2 A – 3 = 0, show that: cos 3 A = 4 cos3 A – 3 cos A
Prove that:
`cosA/(1 + sinA) = secA - tanA`
If `cosec theta = 2x and cot theta = 2/x ," find the value of" 2 ( x^2 - 1/ (x^2))`
Prove the following identity :
`(cosecA - sinA)(secA - cosA) = 1/(tanA + cotA)`
Prove the following identity :
`(sinA - sinB)/(cosA + cosB) + (cosA - cosB)/(sinA + sinB) = 0`
Prove the following identity :
`(cosecθ)/(tanθ + cotθ) = cosθ`
If sec θ + tan θ = m, show that `(m^2 - 1)/(m^2 + 1) = sin theta`
If `1 - cos^2θ = 1/4`, then θ = ?
Prove that `(sec A)/(tan A + cot A) = sin A`.
