If sin-1x-tan-1y then prove that 1x2-1y2-1-

Given,Let #160; #160; sin^ 1x=tan^ 1y=θ #160;or, x=sinθ and y=tanθ or, 1x=cosec θ ......(1) and 1y=θ .....(2).Now, 1x^2 1y^ ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Properties Derived from Trigonometric Identities

If sin-1x=t...

Question

If ${sin}^{- 1} x = {tan}^{- 1} y$ then prove that $\frac{1}{x^{2}} - \frac{1}{y^{2}} = 1$ .

Open in App

Solution

Suggest Corrections

Similar questions

{s i n}^{- 1} x, {c o s}^{- 1} x, {t a n}^{- 1} x

{t a n}^{- 1} x + {t a n}^{- 1} y = {t a n}^{- 1} \frac{x + y}{1 - x y}

x y < 1

π + {t a n}^{- 1} \frac{x + y}{1 - x y}

x y > 1

{c o s}^{- 1} p + {c o s}^{- 1} q + {c o s}^{- 1} r = π

p^{2} + q^{2} + r^{2} + 2 p q r = 1

{s i n}^{- 1} x + {s i n}^{- 1} y + {s i n}^{- 1} z = π

x^{4} + y^{4} + z^{4} + 4 x^{2} y^{2} z^{2} = 2 (x^{2} y^{2} + y^{2} z^{2} + z^{2} x^{2})

{t a n}^{- 1} x + {t a n}^{- 1} y + {t a n}^{- 1} z = π

π / 2

x + y + z = x y z

x y + y z + z x = 1

{sin}^{- 1} x + {sin}^{- 1} y = \frac{π}{2}

\sqrt{1 - x^{2}} \frac{d y}{d x} + \sqrt{1 - y^{2}} = 0

{sin}^{- 1} x, {cos}^{- 1} x, {tan}^{- 1} x

{tan}^{- 1} x + {tan}^{- 1} y = {tan}^{- 1} \frac{x + y}{1 - x y}

x y < 1

π + {tan}^{- 1} \frac{x + y}{1 - x y}

x y > 1

{tan}^{- 1} \frac{1 - x}{1 + x} {tan}^{- 1} \frac{1 - y}{1 + y} = {sin}^{- 1} \frac{y - x}{\sqrt{1 + x^{2}} \sqrt{1 + y^{2}}}

{sin}^{- 1} x + {sin}^{- 1} y = {sin}^{- 1} [x \sqrt{1 - y^{2}} + y \sqrt{1 - x^{2}}]

The general solution of the differential equation $\frac{d y}{d x} = \frac{1 + y^{2}}{1 + x^{2}}$ is

Join BYJU'S Learning Program