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Properties Derived from Trigonometric Identities

If sin-1 x = ...

Question

If $s i n^{- 1} x = θ + β a n d s i n^{- 1} y = θ - β, t h e n 1 + x y =$

A

s i n^{2} θ + s i n^{2} β

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B

s i n^{2} θ + c o s^{2} β

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C

$c o s^{2} θ + c o s^{2} β$

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D

c o s^{2} θ + s i n^{2} β

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Solution

The correct option is B $s i n^{2} θ + c o s^{2} β$
Obvisously $x = s i n (θ + β) a n d y = s i n (θ - β) ∴ 1 + x y = 1 + s i n (θ + β) s i n (θ - β) = 1 + s i n^{2} θ - s i n^{2} β = s i n^{2} θ + c o s^{2} β .$

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Similar questions

(θ - α)

(θ - β)

= y, then prove that

c o s^{2} (α - β) + 2 x y s i n (α - β) = x^{2} + y^{2}

(α, β)

is a point of intersection of the lines

x cos θ + y sin θ = 3 and x sin θ - y cos θ = 4

θ

is parameter, then maximum value of

2^{\frac{α + β}{\sqrt{2}}}

Inverse circular functions,Principal values of

{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

0 \leq x \leq \frac{1}{2}

then the value of

t a n [{s i n}^{- 1} {\frac{x}{\sqrt{2}} + \frac{\sqrt{1 - x^{2}}}{\sqrt{2}}} - {s i n}^{- 1} x]

α = {s i n}^{- 1} \frac{4}{5} + {s i n}^{- 1} \frac{1}{3}

β = {c o s}^{- 1} \frac{4}{5} + {c o s}^{- 1} \frac{1}{3}

cos α + cos β = \frac{3}{2}

sin α + sin β = \frac{1}{2}

θ

is the arithmetic mean of

α

β

sin 2 θ + cos 2 θ

α, β (α > β)

be the roots of the equation

{sin}^{- 1} x - \frac{1}{{sin}^{- 1} x} = {cos}^{- 1} x - \frac{1}{{cos}^{- 1} x} .

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