Advertisements
Advertisements
प्रश्न
If x = asecθ + btanθ and y = atanθ + bsecθ , prove that `x^2 - y^2 = a^2 - b^2`
Advertisements
उत्तर
`x^2 - y^2 = (asecθ + bTanθ)^2 - (aTanθ + bSecθ)^2`
⇒ `a^2sec^2θ + b^2Tan^2θ + 2abSecθTanθ - (a^2Tan^2θ + b^2Sec^2θ + 2abSecθTanθ)`
⇒ `sec^2θ(a^2 - b^2) + Tan^2θ(b^2 - a^2) = (a^2 - b^2)[Sec^2θ - Tan^2θ]`
⇒ `(a^2 - b^2)` [Since `sec^2θ - Tan^2θ = 1`]
Hence , `x^2 - y^2 = a^2 - b^2`
APPEARS IN
संबंधित प्रश्न
Prove the following trigonometric identities.
`(cosec A)/(cosec A - 1) + (cosec A)/(cosec A = 1) = 2 sec^2 A`
If sec θ + tan θ = x, write the value of sec θ − tan θ in terms of x.
Prove the following identity :
`cosA/(1 - tanA) + sinA/(1 - cotA) = sinA + cosA`
Find x , if `cos(2x - 6) = cos^2 30^circ - cos^2 60^circ`
Without using trigonometric identity , show that :
`tan10^circ tan20^circ tan30^circ tan70^circ tan80^circ = 1/sqrt(3)`
If sec θ = `25/7`, then find the value of tan θ.
Prove that `cos θ/sin(90° - θ) + sin θ/cos (90° - θ) = 2`.
Prove that `[(1 + sin theta - cos theta)/(1 + sin theta + cos theta)]^2 = (1 - cos theta)/(1 + cos theta)`
Prove that sin4A – cos4A = 1 – 2 cos2A.
Prove that sec2θ – cos2θ = tan2θ + sin2θ.
