Principal Values MCQ Quiz in मल्याळम - Objective Question with Answer for Principal Values - സൗജന്യ PDF ഡൗൺലോഡ് ചെയ്യുക

Concept:

Calculation:

We have to find the value of sin-1 (0.5)

Let sin-1 (0.5) = θ

⇒ sin θ = 0.5 = \(\frac 1 2\)

⇒ sin θ = sin 30° 

∴ θ =  30° 

Hence θ = sin-1 (0.5)  = 30°

Concept:

• cot^-1 (-x) = π - cot^-1x

• cot (π/3) = 1/√3 

Calculation:

Let y = \(\cot^{-1}\left(-\frac{1}{√{3}}\right)\)

\(⇒ y=π -\cot ^{-1}\frac{1}{√{3}}\)

⇒ y = π - π/3

⇒ y = 2π/3

Since range of cot^-1 is (0, π)

Hence, the principal value is \(\frac{2π}{3}\).

Concept:

cos-1 (-x) = π - cos-1 x, x ∈ [-1, 1]

cos-1 (cos θ) = θ, ∀ θ ∈ [0, π]

Calculation:

As we know that, cos-1 (-x) = π - cos-1 x, x ∈ [-1, 1]

⇒ cos-1 (-1/2) = π - cos-1 (1/2)

⇒ cos-1 (√3/2) + cos-1 (-1/2) = cos-1 (√3/2) + [π - cos-1 (1/2)]

As we know that, cos π/3 = 1/2 and cos π/6 = √3/2

⇒ cos-1 (√3/2) + cos-1 (-1/2) = cos-1 (cos π/6) + [π -  cos-1 (cos π/3)]

As we know that, cos-1 (cos θ) = θ, ∀ θ ∈ [0, π]

⇒ cos-1 (√3/2) + cos-1 (-1/2) = π/6 + [π - π/3] = 5π/6

Concept:

tan (- x) = - tan x

cos (- x) = cos x

sin (- x) = - sin x

tan (2π - x) = - tan x

Calculation: 

Given: \({\tan ^{ - 1}}\left( {\tan \left( {\frac{{7π }}{4}} \right)} \right) \)  we get 

\(\rm \tan \ \frac {7π}{4} = tan\ (2π - \frac {π}{4}) = tan \ (- \frac {π}{4}) = - \tan \ \frac {π}{4} = - \ 1\)

tan^-1 (- 1) = - tan^-1 (1) = \(\rm -\ \frac{\pi}{4}\)

Additional Information

Explanation:

If sinθ = x ⇒ θ = sin^-1x, for θ ∈ [-π/2, π/2]

cot (cot^-1 x) =x for x ∈ R

We have, \(\cot \left[\cos^{−1}\left(\frac{7}{25}\right)\right]\)

Let \(\cos^{−1}\left(\frac{7}{25}\right)\)= θ

⇒ cosθ = 7/25

⇒ sinθ = 24/25

⇒ cotθ = 7/24

∴  \(\cot \left[\cos^{−1}\left(\frac{7}{25}\right)\right]\)

= cotθ 

= 7/24

Concept:

Trigonometric ratios:

• sin θ = \(\frac{perpendicular}{hypotenuse}\)
• cos θ = \(\frac{base}{hypotenuse}\)
• tan θ = \(\frac{perpendicular}{base}\)

Trigonometric Formulae:

• cos 3θ = 4 cos³θ − 3 cos θ
• cos(π − θ) = − cos θ

Calculation:

In anc Δ ABC and Δ PQR, by using trigonometry ratio formula,

cos α = \(\frac{4}{5}\)

Now we know that 

cos 3α = 4 cos³α − 3 cos α

⇒ cos 3α = \(4\times\big(\frac{4}{5}\big)^3-3\times\frac{4}{5} \)

⇒ cos 3α = \(4\times\big(\frac{64}{125}\big)-3\times\frac{4}{5} \)

⇒ cos 3α = \(-\frac{44}{125} \)     ----(1)

Since, cos(π − θ) = − cos θ

⇒ cos(π − 3α) = − cos 3α 

⇒ cos(π − 3α) =  \(\frac{44}{125} \)    [From equation (1)]

​⇒ (π − 3α) = cos^-1 \(\frac{44}{125} \)

∴ The value of  cos-1(\(\frac{44}{125} \)) is (π − 3α).

Concept:

Inverse Trigonometric Functions for Negative Arguments:

Calculation: 

As we know sin-1 (-x) = - sin-1 x

So,  \(\rm \sin^{-1} \left ( \frac {-\ 1}{2}\right) = - \sin^{-1} \left ( \frac {1}{ 2}\right)\)

Let \( \sin^{-1} \left ( \frac {1}{ 2}\right) \) = θ 

⇒ sin θ = \(\frac{1}{ 2}\) = sin 30° 

∴ θ = 30° 

Hence, \(\rm \sin^{-1} \left ( \frac {-1}{2}\right)\) = -θ = -30°

Calculation

Given

tan^-1x + tan^-1 2x = \(\frac{\pi}{4}\); x > 0

⇒ tan-1 2x = \(\frac{\pi}{4}\) - tan-1x 

Taking tan both sides

⇒ 2x = \(\frac{1-x}{1+x}\)

⇒ 2x² + 3x - 1 = 0

⇒ x = \(\frac{-3 \pm \sqrt{9+8}}{8}=\frac{-3 \pm \sqrt{17}}{8}\)

Only possible x = \(\frac{-3+\sqrt{17}}{8}\)

Hence option (2) is correct

We have \(\tan^{-1}\big (\cot \dfrac{3\pi}{4}\big)\)

\(=\tan^{-1}\big(\cot \big(\dfrac{\pi}{2}+\dfrac{\pi}{4}\big)\big)\)

\(=\tan^{-1}(-\tan\dfrac{\pi}{4}\big)\)

\(=\tan^{-1}(\tan(-\dfrac{\pi}{4}\big))\)

\(=-\dfrac{\pi}{4}\)

Given:

\(\sin [2{\cos ^{ - 1}}\frac{{\sqrt 5 }}{3}] \)

Formula Used:

2cos^-1x = cos^-1(2x² - 1)

\(cos^{-1}x = sin^{-1}(\sqrt{1 - x^2})\)

Calculation:

We have,

\(\sin [2{\cos ^{ - 1}}\frac{{\sqrt 5 }}{3}] \)

⇒ \(sin \left[cos^{-1}\left( 2(\frac{\sqrt5}{3})^2 - 1\right) \right]\)

⇒ \(sin\left[ cos^{-1} \left(2 \times \frac{5}{9} - 1 \right) \right]\)

⇒ \(sin \left[ cos^{-1} \frac{1}{9}\right]\)

⇒ \(sin\left[sin^{-1} \sqrt{1 - (\frac{1}{9})^2 } \ \right]\)

⇒ \(sin \left[ sin^{-1} \sqrt{\frac{80}{81}} \right] \)

⇒ \(\frac{\sqrt{80}}{9}\)

⇒ \(\frac{4\sqrt5}{9}\)

∴ The value of \(\sin [2{\cos ^{ - 1}}\frac{{\sqrt 5 }}{3}] \) is \(\frac{4\sqrt5}{9}\)