Advertisements
Advertisements
Question
Prove the following identities:
`1/(sin θ + cos θ) + 1/(sin θ - cos θ) = (2sin θ)/(1 - 2 cos^2 θ)`.
Advertisements
Solution
LHS = `1/(sin θ + cos θ) + 1/(sin θ - cos θ)`
= `((sin θ - cos θ) + (sin θ + cos θ))/(sin^2 θ - cos^2 θ)`
= `(2 sin θ)/((1 - cos^2 θ) - cos^2 θ)`
= `(2 sin θ)/(1 - 2cos^2 θ)`
= RHS
Hence proved.
RELATED QUESTIONS
Prove that sin6θ + cos6θ = 1 – 3 sin2θ. cos2θ.
If `x/a=y/b = z/c` show that `x^3/a^3 + y^3/b^3 + z^3/c^3 = (3xyz)/(abc)`.
Prove the following trigonometric identities
tan2 A + cot2 A = sec2 A cosec2 A − 2
Prove the following identities:
`(secA - tanA)/(secA + tanA) = 1 - 2secAtanA + 2tan^2A`
Prove the following identities:
(sin A + cosec A)2 + (cos A + sec A)2 = 7 + tan2 A + cot2 A
If `cosA/cosB = m` and `cosA/sinB = n`, show that : (m2 + n2) cos2 B = n2.
Prove that: (1+cot A - cosecA)(1 + tan A+ secA) =2.
Prove that cosec2 (90° - θ) + cot2 (90° - θ) = 1 + 2 tan2 θ.
Prove that `costheta/(1 + sintheta) = (1 - sintheta)/(costheta)`
If 4 tanβ = 3, then `(4sinbeta-3cosbeta)/(4sinbeta+3cosbeta)=` ______.
