If cos-13-5-sin-14-5-cos-1x then x equal to -None of these01-1

The correct option is C 1We have cos^ 13/5 sin^ 14/5=cos^ 1x ⇒ sin^ 1[√(1 a/25) ] sin^ 14/5=cos^ 1x ⇒ sin^ 14/5 sin^ 14/5=cos^ 1x ⇒ cos^ 1x=0⇒ x=cos0=1 ∴ x=1 ...

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Byju's Answer

Standard XII

Mathematics

Properties Derived from Trigonometric Identities

If cos^-13/5-...

Question

If $c o s^{- 1} \frac{3}{5} - s i n^{- 1} \frac{4}{5} = c o s^{- 1} x$ then x equal to

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None of these

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Solution

The correct option is A 1
We have
$c o s^{- 1} \frac{3}{5} - s i n^{- 1} \frac{4}{5} = c o s^{- 1} x$
$\Rightarrow s i n^{- 1} [\sqrt{1 - \frac{a}{25}}] - s i n^{- 1} \frac{4}{5} = c o s^{- 1} x$
$\Rightarrow s i n^{- 1} \frac{4}{5} - s i n^{- 1} \frac{4}{5} = c o s^{- 1} x$
$\Rightarrow c o s^{- 1} x = 0 \Rightarrow x = c o s 0 = 1$
$∴ x = 1$

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Similar questions

If $c o s^{- 1} \frac{3}{5} - s i n^{- 1} \frac{4}{5} = c o s^{- 1} x$ then x equal to

Q. If

{c o s}^{- 1} \frac{3}{5} - {s i n}^{- 1} \frac{4}{5} = {c o s}^{- 1} x

, then x is equal to -

{cos}^{- 1} \frac{3}{5} - {sin}^{- 1} \frac{4}{5} = {cos}^{- 1} x

then

x

is equal to:

{cos}^{- 1} (\frac{3}{5}) - {sin}^{- 1} (\frac{4}{5}) = {cos}^{- 1} x

then

x =

Q. 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

.
Solve
(a)

c o s (2 {s i n}^{- 1} x) = 1 / 9

(b)

{c o s}^{- 1} (3 / 5) - {s i n}^{- 1} (4 / 5) = {c o s}^{- 1} x

s i n ({s i n}^{- 1} \frac{1}{5} + {c o s}^{- 1} x) = 1

, then prove that x is equal to

1 / 5

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If cos−135−sin−145=cos−1x then x equal to

The correct option is A 1We have cos−135−sin−145=cos−1x ⇒sin−1[√1−a25]−sin−145=cos−1x ⇒sin−145−sin−145=cos−1x ⇒cos−1x=0⇒x=cos0=1 ∴x=1

If $c o s^{- 1} \frac{3}{5} - s i n^{- 1} \frac{4}{5} = c o s^{- 1} x$ then x equal to