Total number of positive integral value of `n' such that the equations c o s − 1 x + ( s i n − 1 y ) 2 = n π 2 4 and ( s i n − 1 y ) 2 − c o s − 1 x = π 2 16 are consistent, is equal to : Options: A . &nbsp 1 B . &nbsp 4 C . &nbsp 3 D . &nbsp 2

Answer: Option A : A ( s i n − 1 y ) 2 + c o s − 1 x = n π 2 4 ( s i n − 1 y ) 2 − c o s − 1 x = π 2 16 ⇒ ( s i n − 1 y ) 2 = ( 4 n + 1 ) π 2 32 , c o s − 1 x = π 2 ( 4 n − 1 ) 32 ⇒ 0 ≤ ( 4 n + 1 ) 32 π 2 ≤ π 2 4 , 0 ≤ ( 4 n − 1 ) 32 π 2 ≤ π ⇒ − 1 4 ≤ n ≤ 7 4 , 1 4 , ≤ n ≤ 8 π + 1 4

Total number of positive integral value of `n' such that the equations co

Discussion Forum : Inverse Trigonometric Functions

Question -

Total number of positive integral value of `n' such that the equations $c o s^{- 1} x + (s i n^{- 1} y)^{2} = \frac{n π^{2}}{4}$ and $(s i n^{- 1} y)^{2} - c o s^{- 1} x = \frac{π^{2}}{16}$ are consistent, is equal to :

Options:

A . 1

B . 4

C . 3

D . 2

Answer: Option A
:
A

(s i n^{- 1} y)^{2} + c o s^{- 1} x = \frac{n π^{2}}{4}

(s i n^{- 1} y)^{2} - c o s^{- 1} x = \frac{π^{2}}{16}

\Rightarrow (s i n^{- 1} y)^{2} = \frac{(4 n + 1) π^{2}}{32}, c o s^{- 1} x = \frac{π^{2} (4 n - 1)}{32}

\Rightarrow 0 \leq \frac{(4 n + 1)}{32} π^{2} \leq \frac{π^{2}}{4}, 0 \leq \frac{(4 n - 1)}{32} π^{2} \leq π

\Rightarrow - \frac{1}{4} \leq n \leq \frac{7}{4}, \frac{1}{4}, \leq n \leq \frac{8}{π} + \frac{1}{4}

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Discussion Forum : Inverse Trigonometric Functions

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