Let $ \cos (\alpha + \beta ) = $ $ \dfrac{4}{5} $ and let $ \sin (\alpha - \beta ) = $ $ \dfrac{5}{{13}} $ , where 0 ≤ α, β ≤ $ \dfrac{\pi }{4} $ , then $ \tan 2\alpha $ is equal to
A. $ \dfrac{{20}}{7} $
B. \[\dfrac{{25}}{{16}}\]
C. $ \dfrac{{56}}{{33}} $
D. $ \dfrac{{19}}{{12}} $
Answer
584.7k+ views
Hint: We know the cosine of any angle is the ratio of the base to hypotenuse in a right angled triangle. Similarly, sine of any angle is the ratio of the opposite side to hypotenuse in a right angled triangle. Further, we will use Pythagoras theorem to calculate the other trigonometry ratio which is $ \tan 2\alpha $ .
Formula Used: The following formula is used to get to the final answer,
$ \tan (\alpha + \beta ) = $ $ \dfrac{{\tan \alpha + \tan \beta }}{{1 - \tan \alpha \tan \beta }} $
Complete step-by-step answer:
According to the given information, we have
The first function is $ \cos (\alpha + \beta ) = $ $ \dfrac{4}{5} $
Also, we know $ \cos (\alpha + \beta ) = \dfrac{4}{5} $
Therefore take the base to be 4k and hypotenuse as 5k, where k is a constant.
Further, perpendicular = $ \sqrt {{{(hypotenuse)}^2}\;-{\text{ }}{{\left( {base} \right)}^2}} $
$ \Rightarrow \sqrt {{{(5k)}^2} - {{(4k)}^2}} $ $ = \sqrt {9{k^2}} $
Finally we get the perpendicular to be 3k.
Now for a right angled triangle ABC we know that,
$ \tan (\alpha + \beta ) = $ $ \dfrac{3}{4} $ $ ...(1) $
The second function is $ \sin (\alpha - \beta ) = $ $ \dfrac{5}{{13}} $
Also, we know $ \sin (\alpha - \beta ) = \dfrac{5}{{13}} $
Therefore, take the perpendicular to be 5k and hypotenuse as 13k, where k is a constant.
Further, base = $ \sqrt {{{(hypotenuse)}^2}\;-{\text{ }}{{\left( {perpendicular} \right)}^2}} $
$ \Rightarrow \sqrt {{{(13k)}^2} - {{(5k)}^2}} $ $ = \sqrt {144{k^2}} $
Finally we get the base to be 12k.
Now for a right angled triangle ABC we know that,
Hence we get,
$ \tan (\alpha - \beta ) = $ $ \dfrac{5}{{12}} $ $ ...(2) $
According to the given data,
The following formula is used to get,
$ \tan (\alpha + \beta ) = $ $ \dfrac{{\tan \alpha + \tan \beta }}{{1 - \tan \alpha \tan \beta }} $
$ \tan ((\alpha + \beta ) + (\alpha - \beta )) $ = $ \dfrac{{\tan (\alpha + \beta ) + \tan (\alpha - \beta )}}{{1 - \tan (\alpha + \beta )\tan (\alpha - \beta )}} $
On simplifying further we get,
$ \tan 2\alpha $ $ = \dfrac{{\dfrac{3}{4} + \dfrac{5}{{12}}}}{{1 - \dfrac{3}{4} \times \dfrac{5}{{12}}}} $
$ \Rightarrow \dfrac{{\dfrac{{36 + 20}}{{48}}}}{{\dfrac{{48 - 15}}{{48}}}} $ $ = \dfrac{{56}}{{33}} $
So, the correct answer is “$ \dfrac{{56}}{{33}} $ ”.
Note: In order to solve problems of this type the key is to have a basic understanding of trigonometric equations and values and also learn its implications. Students should be aware of applications of the trigonometric values in order to simplify the given equation.
Formula Used: The following formula is used to get to the final answer,
$ \tan (\alpha + \beta ) = $ $ \dfrac{{\tan \alpha + \tan \beta }}{{1 - \tan \alpha \tan \beta }} $
Complete step-by-step answer:
According to the given information, we have
The first function is $ \cos (\alpha + \beta ) = $ $ \dfrac{4}{5} $
Also, we know $ \cos (\alpha + \beta ) = \dfrac{4}{5} $
Therefore take the base to be 4k and hypotenuse as 5k, where k is a constant.
Further, perpendicular = $ \sqrt {{{(hypotenuse)}^2}\;-{\text{ }}{{\left( {base} \right)}^2}} $
$ \Rightarrow \sqrt {{{(5k)}^2} - {{(4k)}^2}} $ $ = \sqrt {9{k^2}} $
Finally we get the perpendicular to be 3k.
Now for a right angled triangle ABC we know that,
$ \tan (\alpha + \beta ) = $ $ \dfrac{3}{4} $ $ ...(1) $
The second function is $ \sin (\alpha - \beta ) = $ $ \dfrac{5}{{13}} $
Also, we know $ \sin (\alpha - \beta ) = \dfrac{5}{{13}} $
Therefore, take the perpendicular to be 5k and hypotenuse as 13k, where k is a constant.
Further, base = $ \sqrt {{{(hypotenuse)}^2}\;-{\text{ }}{{\left( {perpendicular} \right)}^2}} $
$ \Rightarrow \sqrt {{{(13k)}^2} - {{(5k)}^2}} $ $ = \sqrt {144{k^2}} $
Finally we get the base to be 12k.
Now for a right angled triangle ABC we know that,
Hence we get,
$ \tan (\alpha - \beta ) = $ $ \dfrac{5}{{12}} $ $ ...(2) $
According to the given data,
The following formula is used to get,
$ \tan (\alpha + \beta ) = $ $ \dfrac{{\tan \alpha + \tan \beta }}{{1 - \tan \alpha \tan \beta }} $
$ \tan ((\alpha + \beta ) + (\alpha - \beta )) $ = $ \dfrac{{\tan (\alpha + \beta ) + \tan (\alpha - \beta )}}{{1 - \tan (\alpha + \beta )\tan (\alpha - \beta )}} $
On simplifying further we get,
$ \tan 2\alpha $ $ = \dfrac{{\dfrac{3}{4} + \dfrac{5}{{12}}}}{{1 - \dfrac{3}{4} \times \dfrac{5}{{12}}}} $
$ \Rightarrow \dfrac{{\dfrac{{36 + 20}}{{48}}}}{{\dfrac{{48 - 15}}{{48}}}} $ $ = \dfrac{{56}}{{33}} $
So, the correct answer is “$ \dfrac{{56}}{{33}} $ ”.
Note: In order to solve problems of this type the key is to have a basic understanding of trigonometric equations and values and also learn its implications. Students should be aware of applications of the trigonometric values in order to simplify the given equation.
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