The numerical value of tan 2 tan⁻¹...

MathematicsProperties of Inverse Trigonometric FunctionsJEE Advanced 1984Easy

The numerical value of $tan {2 tan^{- 1} (\frac{1}{5}) - \frac{π}{4}}$ is equal to .........

Answer:

- \frac{7}{17}

Visualized Solution (English)

Analyze the Expression

Given expression: $tan {2 tan^{- 1} (\frac{1}{5}) - \frac{π}{4}}$
Objective: Simplify the inner terms into a single $tan^{- 1}$ form.
We will use the identity for $2 tan^{- 1} x$ first.

Apply .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2 tan^{- 1} x Identity

Using the identity: $2 tan^{- 1} x = tan^{- 1} (\frac{2 x}{1 - x ^{2}})$
Substitute $x = \frac{1}{5}$ :
$2 tan^{- 1} (\frac{1}{5}) = tan^{- 1} (\frac{2 ( 1/5 )}{1 - ( 1/5 ) ^{2}})$

Simplify the Fraction

Numerator: $2 \times \frac{1}{5} = \frac{2}{5}$
Denominator: $1 - \frac{1}{25} = \frac{24}{25}$
Result: $tan^{- 1} (\frac{2/5}{24/25}) = tan^{- 1} (\frac{2}{5} \times \frac{25}{24}) = tan^{- 1} (\frac{5}{12})$

Convert .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } \frac{π}{4} to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } tan^{- 1} Form

We know that $tan (\frac{π}{4}) = 1$ , so $\frac{π}{4} = tan^{- 1} (1)$ .
The expression becomes: $tan [tan^{- 1} (\frac{5}{12}) - tan^{- 1} (1)]$

Apply .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } tan^{- 1} A - tan^{- 1} B Identity

Using identity: $tan^{- 1} A - tan^{- 1} B = tan^{- 1} (\frac{A - B}{1 + A B})$
Substitute $A = \frac{5}{12}$ and $B = 1$ :
$tan^{- 1} (\frac{5/12 - 1}{1 + ( 5/12 ) ( 1 )}) = tan^{- 1} (\frac{- 7/12}{17/12}) = tan^{- 1} (- \frac{7}{17})$

Final Evaluation

The expression is now: $tan [tan^{- 1} (- \frac{7}{17})]$
Using the property $tan (tan^{- 1} x) = x$ :
Final Value $= - \frac{7}{17}$

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Conceptually Similar Problems

MathematicsProperties of Inverse Trigonometric FunctionsJEE Advanced 1983Easy

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The value of $tan [cos^{- 1} (\frac{4}{5}) + tan^{- 1} (\frac{2}{3})]$ is

(A)

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(B)

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(C)

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(D)

none

Answer:D

MathematicsProperties of Inverse Trigonometric FunctionsJEE Advanced 1994Easy

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If we consider only the principle values of the inverse trigonometric functions then the value of $tan (cos^{- 1} \frac{1}{5 2} - sin^{- 1} \frac{4}{17})$ is

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(B)

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(C)

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(D)

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Answer:D

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The value of $cot (cosec^{- 1} \frac{5}{3} + tan^{- 1} \frac{2}{3})$ is

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(B)

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Answer:A

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Let $f (θ) = sin (tan^{- 1} (\frac{s i n θ}{c o s 2 θ}))$ , where $- \frac{π}{4} < θ < \frac{π}{4}$ . Then the value of $\frac{d}{d ( t a n θ )} (f (θ))$ is

Answer:1

MathematicsProperties of Inverse Trigonometric FunctionsJEE Advanced 1981Easy

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Find the value of : $cos (2 cos^{- 1} x + sin^{- 1} x)$ at $x = \frac{1}{5}$ , where $0 \leq cos^{- 1} x \leq π$ and $- π /2 \leq sin^{- 1} x \leq π /2$ .

Answer:

- \frac{2 6}{5}

MathematicsFundamental Theorem & Properties of Definite IntegralsJEE Advanced 1993Moderate

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The value of $\int_{π /4}^{3 π /4} \frac{ϕ}{1 + s i n ϕ} d ϕ$ is $\dots\dots\dots$

Answer:

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MathematicsFundamental Theorem & Properties of Definite IntegralsJEE Advanced 1997Moderate

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Determine the value of $\int_{- π}^{π} \frac{2 x ( 1 + s i n x )}{1 + c o s ^{2} x} d x$ .

Answer:

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MathematicsFundamental Theorem & Properties of Definite IntegralsJEE Advanced 2004Moderate

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Find the value of $\int_{- π /3}^{π /3} \frac{π + 4 x ^{3}}{2 - c o s ( ∣ x ∣ + \frac{π}{3} )} d x$ .

Answer:

\frac{4 π}{3} [tan^{- 1} 3 - \frac{π}{4}]

MathematicsFundamental Theorem & Properties of Definite IntegralsJEE Advanced 1993Moderate

View in:English Hindi

The value of $\int_{0}^{π /2} \frac{d x}{1 + t a n ^{3} x}$ is

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MathematicsFundamental Theorem & Properties of Definite IntegralsJEE Advanced 1998Moderate

View in:English Hindi

Prove that $\int_{0}^{1} tan^{- 1} (\frac{1}{1 - x + x ^{2}}) d x = 2 \int_{0}^{1} tan^{- 1} x d x$ . Hence or otherwise, evaluate the integral $\int_{0}^{1} tan^{- 1} (1 - x + x^{2}) d x$ .

Answer:

lo g 2

The numerical value of $tan {2 tan^{- 1} (\frac{1}{5}) - \frac{π}{4}}$ is equal to .........

Visualized Solution (English)

Analyze the Expression

Apply $2 tan^{- 1} x$ Identity

Simplify the Fraction

Convert $\frac{π}{4}$ to $tan^{- 1}$ Form

Apply $tan^{- 1} A - tan^{- 1} B$ Identity

Final Evaluation

Conceptually Similar Problems

The value of $tan [cos^{- 1} (\frac{4}{5}) + tan^{- 1} (\frac{2}{3})]$ is

If we consider only the principle values of the inverse trigonometric functions then the value of $tan (cos^{- 1} \frac{1}{5 2} - sin^{- 1} \frac{4}{17})$ is

The value of $cot (cosec^{- 1} \frac{5}{3} + tan^{- 1} \frac{2}{3})$ is

Let $f (θ) = sin (tan^{- 1} (\frac{s i n θ}{c o s 2 θ}))$ , where $- \frac{π}{4} < θ < \frac{π}{4}$ . Then the value of $\frac{d}{d ( t a n θ )} (f (θ))$ is

Find the value of : $cos (2 cos^{- 1} x + sin^{- 1} x)$ at $x = \frac{1}{5}$ , where $0 \leq cos^{- 1} x \leq π$ and $- π /2 \leq sin^{- 1} x \leq π /2$ .

The value of $\int_{π /4}^{3 π /4} \frac{ϕ}{1 + s i n ϕ} d ϕ$ is $\dots\dots\dots$

Determine the value of $\int_{- π}^{π} \frac{2 x ( 1 + s i n x )}{1 + c o s ^{2} x} d x$ .

Find the value of $\int_{- π /3}^{π /3} \frac{π + 4 x ^{3}}{2 - c o s ( ∣ x ∣ + \frac{π}{3} )} d x$ .

The value of $\int_{0}^{π /2} \frac{d x}{1 + t a n ^{3} x}$ is

Prove that $\int_{0}^{1} tan^{- 1} (\frac{1}{1 - x + x ^{2}}) d x = 2 \int_{0}^{1} tan^{- 1} x d x$ . Hence or otherwise, evaluate the integral $\int_{0}^{1} tan^{- 1} (1 - x + x^{2}) d x$ .

The numerical value of $tan {2 tan^{- 1} (\frac{1}{5}) - \frac{π}{4}}$ is equal to .........

The value of $tan [cos^{- 1} (\frac{4}{5}) + tan^{- 1} (\frac{2}{3})]$ is

If we consider only the principle values of the inverse trigonometric functions then the value of $tan (cos^{- 1} \frac{1}{5 2} - sin^{- 1} \frac{4}{17})$ is

The value of $cot (cosec^{- 1} \frac{5}{3} + tan^{- 1} \frac{2}{3})$ is

Let $f (θ) = sin (tan^{- 1} (\frac{s i n θ}{c o s 2 θ}))$ , where $- \frac{π}{4} < θ < \frac{π}{4}$ . Then the value of $\frac{d}{d ( t a n θ )} (f (θ))$ is

Find the value of : $cos (2 cos^{- 1} x + sin^{- 1} x)$ at $x = \frac{1}{5}$ , where $0 \leq cos^{- 1} x \leq π$ and $- π /2 \leq sin^{- 1} x \leq π /2$ .

The value of $\int_{π /4}^{3 π /4} \frac{ϕ}{1 + s i n ϕ} d ϕ$ is $\dots\dots\dots$

Determine the value of $\int_{- π}^{π} \frac{2 x ( 1 + s i n x )}{1 + c o s ^{2} x} d x$ .

Find the value of $\int_{- π /3}^{π /3} \frac{π + 4 x ^{3}}{2 - c o s ( ∣ x ∣ + \frac{π}{3} )} d x$ .

The value of $\int_{0}^{π /2} \frac{d x}{1 + t a n ^{3} x}$ is

Prove that $\int_{0}^{1} tan^{- 1} (\frac{1}{1 - x + x ^{2}}) d x = 2 \int_{0}^{1} tan^{- 1} x d x$ . Hence or otherwise, evaluate the integral $\int_{0}^{1} tan^{- 1} (1 - x + x^{2}) d x$ .

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The numerical value of tan{2tan−1(51​)−4π​} is equal to .........

Visualized Solution (English)

Analyze the Expression

Apply .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 2tan−1x Identity

Simplify the Fraction

Convert .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 4π​ to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } tan−1 Form

Apply .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } tan−1A−tan−1B Identity

Final Evaluation

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Let $f (θ) = sin (tan^{- 1} (\frac{s i n θ}{c o s 2 θ}))$ , where $- \frac{π}{4} < θ < \frac{π}{4}$ . Then the value of $\frac{d}{d ( t a n θ )} (f (θ))$ is

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The value of $cot (cosec^{- 1} \frac{5}{3} + tan^{- 1} \frac{2}{3})$ is

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