If 2 x 3 4 , then cos – 1 ( 12 13 cos x + 5 13 sin x ) is equal to [2025]

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Q.
If $\frac{π}{2} \leq x \leq \frac{3 π}{4}$ , then $\cos^{- 1} (\frac{12}{13} \cos x + \frac{5}{13} \sin x)$ is equal to [2025]

1 $x - \tan^{- 1} \frac{5}{12}$

2 $x + \tan^{- 1} \frac{5}{12}$

3 $x + \tan^{- 1} \frac{4}{5}$

4 $x - \tan^{- 1} \frac{4}{3}$

Ans.
(1)

Given, $\frac{π}{2} \leq x \leq \frac{3 π}{4}$

Let $\cos α = \frac{12}{13} [∴ \cos^{- 1} (\frac{12}{13} \cos x + \frac{5}{13} \sin x)]$

$= \cos^{- 1} (\cos x \cos α + \sin x \sin α) = \cos^{- 1} (\cos (x - α)) = x - α$

$= x - \tan^{- 1} (\frac{5}{12})$

Solution

Q.
If $\frac{π}{2} \leq x \leq \frac{3 π}{4}$ , then $\cos^{- 1} (\frac{12}{13} \cos x + \frac{5}{13} \sin x)$ is equal to [2025]

1 $x - \tan^{- 1} \frac{5}{12}$

2 $x + \tan^{- 1} \frac{5}{12}$

3 $x + \tan^{- 1} \frac{4}{5}$

4 $x - \tan^{- 1} \frac{4}{3}$

Ans.
(1)

Given, $\frac{π}{2} \leq x \leq \frac{3 π}{4}$

Let $\cos α = \frac{12}{13} [∴ \cos^{- 1} (\frac{12}{13} \cos x + \frac{5}{13} \sin x)]$

$= \cos^{- 1} (\cos x \cos α + \sin x \sin α) = \cos^{- 1} (\cos (x - α)) = x - α$

$= x - \tan^{- 1} (\frac{5}{12})$

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Solution

Q. If π2≤x≤3π4, then cos–1(1213cosx+513sinx) is equal to [2025]

1 x–tan–1512

2 x+tan–1512

3 x+tan–145

4 x–tan–143

Ans. (1) Given, π2≤x≤3π4 Let cosα=1213 [∴ cos–1(1213cosx+513sinx)] =cos–1(cosx cosα+sinx sinα)=cos–1(cos(x–α))=x–α =x–tan–1(512)

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Q.
If $\frac{π}{2} \leq x \leq \frac{3 π}{4}$ , then $\cos^{- 1} (\frac{12}{13} \cos x + \frac{5}{13} \sin x)$ is equal to [2025]

1 $x - \tan^{- 1} \frac{5}{12}$

2 $x + \tan^{- 1} \frac{5}{12}$

3 $x + \tan^{- 1} \frac{4}{5}$

4 $x - \tan^{- 1} \frac{4}{3}$

Ans.
(1)

Given, $\frac{π}{2} \leq x \leq \frac{3 π}{4}$

Let $\cos α = \frac{12}{13} [∴ \cos^{- 1} (\frac{12}{13} \cos x + \frac{5}{13} \sin x)]$

$= \cos^{- 1} (\cos x \cos α + \sin x \sin α) = \cos^{- 1} (\cos (x - α)) = x - α$

$= x - \tan^{- 1} (\frac{5}{12})$