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Question : 17
Total: 29
Prove that: s i n − 1 (
) + s i n − 1 (
) + s i n − 1 (
) =
OR
Solve for x:t a n − 1 3 x + t a n − 1 2 x =
OR
Solve for x:
Solution:
To prove: s i n − 1 (
) + s i n − 1 (
) + s i n − 1 (
) =
Lets i n − 1 (
) = x
⇒ sin x =
⇒ cos x =√ 1 − s i n 2 x =
s i n − 1 (
) = y
⇒ sin y =
⇒ cos y =√ 1 − s i n 2 y =
s i n − 1 (
) = z
⇒ sin z =
⇒ cos z =√ 1 − s i n 2 x =
tan x =
, tan y =
, tan z =
tan z =
⇒ cot z =
... (1)
tan (x + y) =
⇒ tan (x + y) =
⇒ tan (x + y) =
⇒ tan (x + y) = cot z ... [from equation (1)]
⇒ tan (x + y) = tan(
− z )
⇒ x + y =
- z
⇒ x + y + z =
∴s i n − 1 (
) + s i n − 1 (
) + s i n − 1 (
) =
OR
t a n − 1 3 x + t a n − 1 2 x =
⇒t a n − 1 (
) =
. 3x × 2x < 1
⇒ tan[ t a n − 1 (
) ] = tan
⇒
= 1
⇒ 1 -6 x 2 = 5x
⇒6 x 2 + 5x - 1 = 0
⇒6 x 2 + 6x - x - 1 = 0
⇒ x = - 1 or
Here (- 3) × (- 2) ≮ 1 [Since (- 3) × (- 2) = 6 >1]
Therefore, x = 1 is not the solution.
When substituting x =
in 3x 2x, we have,
3 ×
× 2 ×
=
×
=
< 1
Hence x =
is the solution of the given equation
Let
⇒ sin x =
⇒ cos x =
⇒ sin y =
⇒ cos y =
⇒ sin z =
⇒ cos z =
tan x =
tan z =
tan (x + y) =
⇒ tan (x + y) =
⇒ tan (x + y) =
⇒ tan (x + y) = cot z ... [from equation (1)]
⇒ tan (x + y) = tan
⇒ x + y =
⇒ x + y + z =
∴
OR
⇒
⇒ tan
⇒
⇒ 1 -
⇒
⇒
⇒ x = - 1 or
Here (- 3) × (- 2) ≮ 1 [Since (- 3) × (- 2) = 6 >1]
Therefore, x = 1 is not the solution.
When substituting x =
3 ×
Hence x =
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