If sin-1 x = tan-1 y, what is the value of (1/x2) - (1/y2) ?

Question

If sin-1 x = tan-1 y, what is the value of (1/x2) - (1/y2) ? (A) 1 (B) -1 (C) 0 (D) 2

Tardigrade Admin · Accepted Answer

Let, sin-1 x = tan-1 y = θ ⇒ x = sinθ y = tanθ (1/x2) = (1/ sin2 θ) = cosec 2 θ and (1/y2) = (1/ tan2 θ) = cot2 θ . ⇒ (1/x2 ) - (1/y2) = cosec 2 θ - cot2 θ = 1

Q. If $sin^{- 1} x = tan^{- 1} y,$ what is the value of $\frac{1}{x ^{2}} - \frac{1}{y ^{2}}$ ?

1

-1

0

2

Let, $sin^{- 1} x = tan^{- 1} y = θ$
$\Rightarrow x = sin θ y = tan θ$
$\frac{1}{x ^{2}} = \frac{1}{s i n ^{2} θ} = cose c^{2} θ$
and $\frac{1}{y ^{2}} = \frac{1}{t a n ^{2} θ} = cot^{2} θ .$
$\Rightarrow \frac{1}{x ^{2}} - \frac{1}{y ^{2}} = cose c^{2} θ - cot^{2} θ = 1$

Q. If sin−1x=tan−1y, what is the value of x21​−y21​ ?

1

-1

0

2

Let, sin−1x=tan−1y=θ ⇒x=sinθy=tanθ x21​=sin2θ1​=cosec2θ and y21​=tan2θ1​=cot2θ. ⇒x21​−y21​=cosec2θ−cot2θ=1

Q. If $sin^{- 1} x = tan^{- 1} y,$ what is the value of $\frac{1}{x ^{2}} - \frac{1}{y ^{2}}$ ?

Let, $sin^{- 1} x = tan^{- 1} y = θ$
$\Rightarrow x = sin θ y = tan θ$
$\frac{1}{x ^{2}} = \frac{1}{s i n ^{2} θ} = cose c^{2} θ$
and $\frac{1}{y ^{2}} = \frac{1}{t a n ^{2} θ} = cot^{2} θ .$
$\Rightarrow \frac{1}{x ^{2}} - \frac{1}{y ^{2}} = cose c^{2} θ - cot^{2} θ = 1$