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 that:
sec2θ + cosec2θ = sec2θ x cosec2θ
Prove that `(sec theta - 1)/(sec theta + 1) = ((sin theta)/(1 + cos theta))^2`
Prove the following identities:
(1 + cot A – cosec A)(1 + tan A + sec A) = 2
Show that : `sinA/sin(90^circ - A) + cosA/cos(90^circ - A) = sec A cosec A`
Write the value of `cosec^2 theta (1+ cos theta ) (1- cos theta).`
Write True' or False' and justify your answer the following :
The value of \[\cos^2 23 - \sin^2 67\] is positive .
Prove the following identity :
`(1 - cos^2θ)sec^2θ = tan^2θ`
Prove the following identity :
`(secθ - tanθ)^2 = (1 - sinθ)/(1 + sinθ)`
Without using trigonometric table , evaluate :
`(sin47^circ/cos43^circ)^2 - 4cos^2 45^circ + (cos43^circ/sin47^circ)^2`
Prove that sec θ. cosec (90° - θ) - tan θ. cot( 90° - θ ) = 1.
