Basic trigo - VI

Geometry Level 2

Let f ( θ ) = cot ⁡ θ 1 + cot ⁡ θ f\left( \theta \right) =\frac { \cot { \theta } }{ 1+\cot { \theta } } and α + β = 5 π 4 \alpha +\beta =\frac { 5\pi }{ 4 } . Then the value of f ( α ) . f ( β ) f\left( \alpha \right) .f\left( \beta \right) can be written as a b \frac { a }{ b } where a a and b b are coprime integers. Find a + b a+b

This question is a part of the set: Basic Trigo


The answer is 3.

Krishna Ramesh by Krishna Ramesh , 22, India

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1 solution

Krishna Ramesh
Jun 14, 2014

f ( α ) . f ( β ) = ( cot ⁡ α 1 + cot ⁡ α ) ( cot ⁡ β 1 + cot ⁡ β ) = cot ⁡ α cot ⁡ β 1 + cot ⁡ α + cot ⁡ β + cot ⁡ α cot ⁡ β f\left( \alpha \right) .f\left( \beta \right) =\left( \frac { \cot { \alpha } }{ 1+\cot { \alpha } } \right) \left( \frac { \cot { \beta } }{ 1+\cot { \beta } } \right) \\ \quad \quad \quad \quad \quad \quad \quad =\frac { \cot { \alpha } \cot { \beta } }{ 1+\cot { \alpha } +\cot { \beta } +\cot { \alpha } \cot { \beta } } \\

now, cot ⁡ ( α + β ) = cot ⁡ α cot ⁡ β − 1 cot ⁡ α + cot ⁡ β ⇒ cot ⁡ ( 5 π 4 ) = cot ⁡ α cot ⁡ β − 1 cot ⁡ α + cot ⁡ β ⇒ 1 = cot ⁡ α cot ⁡ β − 1 cot ⁡ α + cot ⁡ β ⇒ cot ⁡ α cot ⁡ β = cot ⁡ α + cot ⁡ β + 1 \cot { \left( \alpha +\beta \right) } =\frac { \cot { \alpha } \cot { \beta } -1 }{ \cot { \alpha } +\cot { \beta } } \\ \Rightarrow \cot { \left( \frac { 5\pi }{ 4 } \right) } =\frac { \cot { \alpha } \cot { \beta } -1 }{ \cot { \alpha } +\cot { \beta } } \\ \Rightarrow 1=\frac { \cot { \alpha } \cot { \beta } -1 }{ \cot { \alpha } +\cot { \beta } } \Rightarrow \cot { \alpha } \cot { \beta } =\cot { \alpha } +\cot { \beta } +1

so, f ( α ) . f ( β ) = cot ⁡ α cot ⁡ β cot ⁡ α cot ⁡ β + cot ⁡ α cot ⁡ β = 1 2 f\left( \alpha \right) .f\left( \beta \right) =\frac { \cot { \alpha } \cot { \beta } }{ \cot { \alpha } \cot { \beta } +\cot { \alpha } \cot { \beta } } =\frac { 1 }{ 2 }

so, the answer =1+2=3

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I also did the same way!

Kartik Sharma - 6 years, 9 months ago

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Who asked you :P

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