Prove that cos tan⁻¹ sin cot⁻¹ x =...

MathematicsProperties of Inverse Trigonometric FunctionsJEE Advanced 2002Easy

Prove that $cos tan^{- 1} sin cot^{- 1} x = \frac{x ^{2} + 1}{x ^{2} + 2}$ .

Visualized Solution (Hindi)

Analyze the Nested Expression

Given expression: $cos (tan^{- 1} (sin (cot^{- 1} x)))$

Substitute the Innermost Angle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } θ

Let $θ = cot^{- 1} x$
This implies $cot θ = x$

Visualize the First Triangle

In a right triangle with angle $θ$ :
Base $= x$ , Perpendicular $= 1$
Hypotenuse $= x^{2} + 1$

Find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sin θ

$sin θ = \frac{Perpendicular}{Hypotenuse}$
$sin (cot^{- 1} x) = \frac{1}{x ^{2} + 1}$

Update the Main Expression

Expression becomes: $cos (tan^{- 1} (\frac{1}{x ^{2} + 1}))$

Substitute the Next Angle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ϕ

Let $ϕ = tan^{- 1} (\frac{1}{x ^{2} + 1})$
This implies $tan ϕ = \frac{1}{x ^{2} + 1}$

Visualize the Second Triangle

In a new right triangle with angle $ϕ$ :
Base $= x^{2} + 1$ , Perpendicular $= 1$
Hypotenuse $= (x^{2} + 1)^{2} + 1^{2} = x^{2} + 2$

Find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } cos ϕ

$cos ϕ = \frac{Base}{Hypotenuse}$
$cos ϕ = \frac{x ^{2} + 1}{x ^{2} + 2}$

The Final Result

Therefore, $cos (tan^{- 1} (sin (cot^{- 1} x))) = \frac{x ^{2} + 1}{x ^{2} + 2}$
Hence Proved.

Key Takeaway & Next Challenges

Key Takeaway: Use the right-triangle method layer-by-layer for nested ITF problems.
Next Challenge: Try evaluating $sin (tan^{- 1} (cos (cot^{- 1} x)))$ using the same logic.

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Conceptually Similar Problems

MathematicsProperties of Inverse Trigonometric FunctionsJEE Advanced 2008Moderate

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$1 + x^{2} [{x cos (cot^{- 1} x) + sin (cot^{- 1} x)}^{2} - 1]^{1/2} =$

(A)

\frac{x}{1 + x ^{2}}

(B)

(C)

1 + x^{2}

(D)

1 + x^{2}

Answer:C

MathematicsTrigonometric Ratios and IdentitiesJEE Advanced 1988Easy

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Prove that $tan α + 2 tan 2 α + 4 tan 4 α + 8 cot 8 α = cot α$ .

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$cot^{- 1} (cos α) - tan^{- 1} (cos α) = x$ , then $sin x =$

(A)

tan^{2} (\frac{α}{2})

(B)

cot^{2} (\frac{α}{2})

(C)

tan α

(D)

cot (\frac{α}{2})

Answer:A

MathematicsFundamental Theorem & Properties of Definite IntegralsJEE Advanced 1998Moderate

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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

MathematicsConditional IdentitiesJEE Advanced 2000Easy

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In any triangle $A B C$ , prove that $cot \frac{A}{2} + cot \frac{B}{2} + cot \frac{C}{2} = cot \frac{A}{2} cot \frac{B}{2} cot \frac{C}{2}$ .

MathematicsIntegration by SubstitutionJEE Advanced 1989Moderate

View in:English Hindi

Evaluate $\int (tan x + cot x) d x$

Answer:

2 tan^{- 1} (\frac{t a n x - c o t x}{2}) + c

MathematicsIntegration by SubstitutionJEE Advanced 1985Moderate

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Evaluate the following $\int \frac{1 - x}{1 + x} d x$

Answer:

- 2 1 - x + cos^{- 1} x + x 1 - x + C

MathematicsProperties of Inverse Trigonometric FunctionsJEE Advanced 2001Moderate

View in:English Hindi

If $sin^{- 1} (x - \frac{x ^{2}}{2} + \frac{x ^{3}}{4} - \dots) + cos^{- 1} (x^{2} - \frac{x ^{4}}{2} + \frac{x ^{6}}{4} - \dots) = \frac{π}{2}$ for $0 < ∣ x ∣ < 2$ , then $x$ equals

(A)

1/2

(B)

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(D)

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Answer:B

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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MathematicsProperties of Inverse Trigonometric FunctionsJEE Advanced 1993Moderate

View in:English Hindi

Prove that $cos tan^{- 1} sin cot^{- 1} x = \frac{x ^{2} + 1}{x ^{2} + 2}$ .

Visualized Solution (Hindi)

Analyze the Nested Expression

Substitute the Innermost Angle $θ$

Visualize the First Triangle

Find $sin θ$

Update the Main Expression

Substitute the Next Angle $ϕ$

Visualize the Second Triangle

Find $cos ϕ$

The Final Result

Key Takeaway & Next Challenges

Conceptually Similar Problems

$1 + x^{2} [{x cos (cot^{- 1} x) + sin (cot^{- 1} x)}^{2} - 1]^{1/2} =$

Prove that $tan α + 2 tan 2 α + 4 tan 4 α + 8 cot 8 α = cot α$ .

$cot^{- 1} (cos α) - tan^{- 1} (cos α) = x$ , then $sin x =$

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 $\int_{0}^{1} tan^{- 1} (1 - x + x^{2}) d x$ .

In any triangle $A B C$ , prove that $cot \frac{A}{2} + cot \frac{B}{2} + cot \frac{C}{2} = cot \frac{A}{2} cot \frac{B}{2} cot \frac{C}{2}$ .

Evaluate $\int (tan x + cot x) d x$

Evaluate the following $\int \frac{1 - x}{1 + x} d x$

If $sin^{- 1} (x - \frac{x ^{2}}{2} + \frac{x ^{3}}{4} - \dots) + cos^{- 1} (x^{2} - \frac{x ^{4}}{2} + \frac{x ^{6}}{4} - \dots) = \frac{π}{2}$ for $0 < ∣ x ∣ < 2$ , then $x$ equals

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

Using mathematical induction, prove that $tan^{- 1} (1/3) + tan^{- 1} (1/7) + .... tan^{- 1} {1/ (n^{2} + n + 1)} = tan^{- 1} {n / (n + 2)}$

Prove that $cos tan^{- 1} sin cot^{- 1} x = \frac{x ^{2} + 1}{x ^{2} + 2}$ .

$1 + x^{2} [{x cos (cot^{- 1} x) + sin (cot^{- 1} x)}^{2} - 1]^{1/2} =$

Prove that $tan α + 2 tan 2 α + 4 tan 4 α + 8 cot 8 α = cot α$ .

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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 $\int_{0}^{1} tan^{- 1} (1 - x + x^{2}) d x$ .

In any triangle $A B C$ , prove that $cot \frac{A}{2} + cot \frac{B}{2} + cot \frac{C}{2} = cot \frac{A}{2} cot \frac{B}{2} cot \frac{C}{2}$ .

Evaluate $\int (tan x + cot x) d x$

Evaluate the following $\int \frac{1 - x}{1 + x} d x$

If $sin^{- 1} (x - \frac{x ^{2}}{2} + \frac{x ^{3}}{4} - \dots) + cos^{- 1} (x^{2} - \frac{x ^{4}}{2} + \frac{x ^{6}}{4} - \dots) = \frac{π}{2}$ for $0 < ∣ x ∣ < 2$ , then $x$ equals

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

Using mathematical induction, prove that $tan^{- 1} (1/3) + tan^{- 1} (1/7) + .... tan^{- 1} {1/ (n^{2} + n + 1)} = tan^{- 1} {n / (n + 2)}$

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Prove that costan−1sincot−1x=x2+2x2+1​​.

Visualized Solution (Hindi)

Analyze the Nested Expression

Substitute the Innermost Angle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } θ

Visualize the First Triangle

Find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } sinθ

Update the Main Expression

Substitute the Next Angle .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ϕ

Visualize the Second Triangle

Find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } cosϕ

The Final Result

Key Takeaway & Next Challenges

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Substitute the Next Angle $ϕ$

Find $cos ϕ$

$1 + x^{2} [{x cos (cot^{- 1} x) + sin (cot^{- 1} x)}^{2} - 1]^{1/2} =$

Prove that $tan α + 2 tan 2 α + 4 tan 4 α + 8 cot 8 α = cot α$ .

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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 $\int_{0}^{1} tan^{- 1} (1 - x + x^{2}) d x$ .

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If $sin^{- 1} (x - \frac{x ^{2}}{2} + \frac{x ^{3}}{4} - \dots) + cos^{- 1} (x^{2} - \frac{x ^{4}}{2} + \frac{x ^{6}}{4} - \dots) = \frac{π}{2}$ for $0 < ∣ x ∣ < 2$ , then $x$ equals

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Prove that $cos tan^{- 1} sin cot^{- 1} x = \frac{x ^{2} + 1}{x ^{2} + 2}$ .

$1 + x^{2} [{x cos (cot^{- 1} x) + sin (cot^{- 1} x)}^{2} - 1]^{1/2} =$

Prove that $tan α + 2 tan 2 α + 4 tan 4 α + 8 cot 8 α = cot α$ .

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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 $\int_{0}^{1} tan^{- 1} (1 - x + x^{2}) d x$ .

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