If $\tan A = \sqrt 2 - 1$ then $\sin A\cos A = \dfrac{{\sqrt 2 }}{m}$, find the value of m
Answer
Verified360.3k+ views
Hint – In this problem the value of tan A is given to us and we need to find the value of m if sinAcosA is given to us. Use trigonometric ratios of tan A to obtain the value of sinA and then the value of cosA using direct trigonometric ratios. Then multiply the values to obtain the answer.
“Complete step-by-step answer:”
Given data
$\tan A = \sqrt 2 - 1$
And if $\sin A\cos A = \dfrac{{\sqrt 2 }}{m}$.
Then we have to find out the value of m.
$ \Rightarrow \tan A = \dfrac{{\sqrt 2 - 1}}{1}$
As we know tan is the ratio of perpendicular (P) to base (B).
$ \Rightarrow \left( P \right) = \sqrt 2 - 1,{\text{ }}\left( B \right) = 1$
So first calculate the value of hypotenuse (H) by using Pythagoras Theorem we have
${\left( {{\text{Hypotenuse}}} \right)^2} = {\left( {{\text{Perpendicular}}} \right)^2} + {\left( {{\text{Base}}} \right)^2}$
$ \Rightarrow {\left( {\text{H}} \right)^2} = {\left( {\sqrt 2 - 1} \right)^2} + {1^2}$
Now open the whole square using ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$
$ \Rightarrow {\left( {\text{H}} \right)^2} = 2 + 1 - 2\sqrt 2 + 1 = 4 - 2\sqrt 2 $
Now take square root on both sides we have,
$ \Rightarrow H = \sqrt {4 - 2\sqrt 2 } $
Now as we know sine is the ratio of perpendicular to hypotenuse and cosine is the ratio of base to hypotenuse.
$ \Rightarrow \sin A = \dfrac{P}{H} = \dfrac{{\sqrt 2 - 1}}{{\sqrt {4 - 2\sqrt 2 } }},{\text{ }}\cos A = \dfrac{B}{H} = \dfrac{1}{{\sqrt {4 - 2\sqrt 2 } }}$
$ \Rightarrow \sin A\cos A = \dfrac{{\sqrt 2 - 1}}{{4 - 2\sqrt 2 }}$
Now multiply and divide by $\sqrt 2 $ we have,
$ \Rightarrow \sin A\cos A = \dfrac{{\left( {\sqrt 2 - 1} \right)\sqrt 2 }}{{\left( {4 - 2\sqrt 2 } \right)\sqrt 2 }} = \dfrac{{2 - \sqrt 2 }}{{2\sqrt 2 \left( {2 - \sqrt 2 } \right)}} = \dfrac{1}{{2\sqrt 2 }}$
Now again multiply and divide by $\sqrt 2 $ we have,
$ \Rightarrow \sin A\cos A = \dfrac{1}{{2\sqrt 2 }} \times \dfrac{{\sqrt 2 }}{{\sqrt 2 }} = \dfrac{{\sqrt 2 }}{4}$
Now compare this value with the given value we have,
$ \Rightarrow m = 4$.
So, this is the required value of m.
Hence, this is the required answer.
Note – Whenever we face such types of problems the key concept is simply to have a gist of all the basic trigonometric formulas, this will help in direct evaluation of other trigonometric quantities using the given trigonometric entity. This simplification according to the required conditions of problems will get you on the right track to reach the answer.
“Complete step-by-step answer:”
Given data
$\tan A = \sqrt 2 - 1$
And if $\sin A\cos A = \dfrac{{\sqrt 2 }}{m}$.
Then we have to find out the value of m.
$ \Rightarrow \tan A = \dfrac{{\sqrt 2 - 1}}{1}$
As we know tan is the ratio of perpendicular (P) to base (B).
$ \Rightarrow \left( P \right) = \sqrt 2 - 1,{\text{ }}\left( B \right) = 1$
So first calculate the value of hypotenuse (H) by using Pythagoras Theorem we have
${\left( {{\text{Hypotenuse}}} \right)^2} = {\left( {{\text{Perpendicular}}} \right)^2} + {\left( {{\text{Base}}} \right)^2}$
$ \Rightarrow {\left( {\text{H}} \right)^2} = {\left( {\sqrt 2 - 1} \right)^2} + {1^2}$
Now open the whole square using ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$
$ \Rightarrow {\left( {\text{H}} \right)^2} = 2 + 1 - 2\sqrt 2 + 1 = 4 - 2\sqrt 2 $
Now take square root on both sides we have,
$ \Rightarrow H = \sqrt {4 - 2\sqrt 2 } $
Now as we know sine is the ratio of perpendicular to hypotenuse and cosine is the ratio of base to hypotenuse.
$ \Rightarrow \sin A = \dfrac{P}{H} = \dfrac{{\sqrt 2 - 1}}{{\sqrt {4 - 2\sqrt 2 } }},{\text{ }}\cos A = \dfrac{B}{H} = \dfrac{1}{{\sqrt {4 - 2\sqrt 2 } }}$
$ \Rightarrow \sin A\cos A = \dfrac{{\sqrt 2 - 1}}{{4 - 2\sqrt 2 }}$
Now multiply and divide by $\sqrt 2 $ we have,
$ \Rightarrow \sin A\cos A = \dfrac{{\left( {\sqrt 2 - 1} \right)\sqrt 2 }}{{\left( {4 - 2\sqrt 2 } \right)\sqrt 2 }} = \dfrac{{2 - \sqrt 2 }}{{2\sqrt 2 \left( {2 - \sqrt 2 } \right)}} = \dfrac{1}{{2\sqrt 2 }}$
Now again multiply and divide by $\sqrt 2 $ we have,
$ \Rightarrow \sin A\cos A = \dfrac{1}{{2\sqrt 2 }} \times \dfrac{{\sqrt 2 }}{{\sqrt 2 }} = \dfrac{{\sqrt 2 }}{4}$
Now compare this value with the given value we have,
$ \Rightarrow m = 4$.
So, this is the required value of m.
Hence, this is the required answer.
Note – Whenever we face such types of problems the key concept is simply to have a gist of all the basic trigonometric formulas, this will help in direct evaluation of other trigonometric quantities using the given trigonometric entity. This simplification according to the required conditions of problems will get you on the right track to reach the answer.
Last updated date: 22nd Sep 2023
•
Total views: 360.3k
•
Views today: 7.60k
