Advertisements
Advertisements
प्रश्न
Prove the following identity :
`(1 - sin^2θ)sec^2θ = 1`
Advertisements
उत्तर
`(1 - sin^2θ)sec^2θ = 1`
Consider L.H.S = `cos^2θsec^2θ`
= `cos^2θ xx 1/cos^2θ = 1`
= R.H.S
Hence proved.
APPEARS IN
संबंधित प्रश्न
If secθ + tanθ = p, show that `(p^{2}-1)/(p^{2}+1)=\sin \theta`
Prove the following trigonometric identities.
sec A (1 − sin A) (sec A + tan A) = 1
If sin2 θ cos2 θ (1 + tan2 θ) (1 + cot2 θ) = λ, then find the value of λ.
Without using trigonometric identity , show that :
`tan10^circ tan20^circ tan30^circ tan70^circ tan80^circ = 1/sqrt(3)`
Without using trigonometric identity , show that :
`cos^2 25^circ + cos^2 65^circ = 1`
Prove that:
`sqrt(( secθ - 1)/(secθ + 1)) + sqrt((secθ + 1)/(secθ - 1)) = 2 "cosec"θ`
If tan A + sin A = m and tan A − sin A = n, then show that `m^2 - n^2 = 4 sqrt (mn)`.
a cot θ + b cosec θ = p and b cot θ + a cosec θ = q then p2 – q2 is equal to
Prove that cosec θ – cot θ = `(sin θ)/(1 + cos θ)`.
`1/sin^2θ - 1/cos^2θ - 1/tan^2θ - 1/cot^2θ - 1/sec^2θ - 1/("cosec"^2θ) = -3`, then find the value of θ.
