Advertisements
Advertisements
प्रश्न
Prove the following identities:
`tan^2A - tan^2B = (sin^2A - sin^2B)/(cos^2A * cos^2B)`
Advertisements
उत्तर
L.H.S. = `tan^2A - tan^2B`
= `sin^2A/cos^2A - sin^2B/cos^2B`
= `(sin^2A * cos^2B - sin^2B * cos^2A)/(cos^2A * cos^2B`
= `(sin^2A(1 - sin^2B)-sin^2B(1 - sin^2A))/(cos^2A * cos^2B)`
= `(sin^2A - sin^2A * sin^2B - sin^2B + sin^2A * sin^2B)/(cos^2A * cos^2B`
= `(sin^2A - cancel(sin^2A * sin^2B) - sin^2B + cancel(sin^2A * sin^2B))/(cos^2A * cos^2B`
= `(sin^2A - sin^2B)/(cos^2A * cos^2B)` = R.H.S.
APPEARS IN
संबंधित प्रश्न
If acosθ – bsinθ = c, prove that asinθ + bcosθ = `\pm \sqrt{a^{2}+b^{2}-c^{2}`
Prove that `\frac{\sin \theta -\cos \theta }{\sin \theta +\cos \theta }+\frac{\sin\theta +\cos \theta }{\sin \theta -\cos \theta }=\frac{2}{2\sin^{2}\theta -1}`
Prove the following trigonometric identities.
`tan theta - cot theta = (2 sin^2 theta - 1)/(sin theta cos theta)`
If `cos theta = 2/3 , "write the value of" ((sec theta -1))/((sec theta +1))`
\[\frac{x^2 - 1}{2x}\] is equal to
Find the value of `θ(0^circ < θ < 90^circ)` if :
`tan35^circ cot(90^circ - θ) = 1`
If `tan θ = 9/40`, complete the activity to find the value of sec θ.
Activity:
sec2θ = 1 + `square` ...[Fundamental trigonometric identity]
sec2θ = 1 + `square^2`
sec2θ = 1 + `square`
sec θ = `square`
Prove that `(1 + sin θ)/(1 - sin θ) = (sec θ + tan θ)^2`.
If 1 + sin2θ = 3 sin θ cos θ, then prove that tan θ = 1 or `1/2`.
Let x1, x2, x3 be the solutions of `tan^-1((2x + 1)/(x + 1)) + tan^-1((2x - 1)/(x - 1))` = 2tan–1(x + 1) where x1 < x2 < x3 then 2x1 + x2 + x32 is equal to ______.
