
If $\sec \alpha = \dfrac{5}{4}$ verify that $\dfrac{{\tan \alpha }}{{1 + {{\left( {\tan \alpha } \right)}^2}}} = \dfrac{{\sin \alpha }}{{\sec \alpha }}$.
Answer
521.4k+ views
Hint: Here, we have to simplify the L.H.S and R.H.S and convert them into a simplest trigonometric function by using identity $1 + {\left( {\tan \alpha } \right)^2} = {\left( {\sec \alpha } \right)^2}$ .
Complete step-by-step answer:
Given, $\sec \alpha = \dfrac{5}{4}$
To prove: $\dfrac{{\tan \alpha }}{{1 + {{\left( {\tan \alpha } \right)}^2}}} = \dfrac{{\sin \alpha }}{{\sec \alpha }}$
Taking LHS of the above equation which needs to be proved
i.e., ${\text{LHS}} = \dfrac{{\tan \alpha }}{{1 + {{\left( {\tan \alpha } \right)}^2}}}$
Using the identity $1 + {\left( {\tan \alpha } \right)^2} = {\left( {\sec \alpha } \right)^2}$ , we get
${\text{LHS}} = \dfrac{{\tan \alpha }}{{{{\left( {\sec \alpha } \right)}^2}}}{\text{ }} \to {\text{(1)}}$
As we know that $\tan \alpha = \dfrac{{\sin \alpha }}{{\cos \alpha }}$ and $\sec \alpha = \dfrac{1}{{\cos \alpha }}$ so we can write that $\tan \alpha = \left( {\sin \alpha } \right)\left( {\sec \alpha } \right)$
Now put $\tan \alpha = \left( {\sin \alpha } \right)\left( {\sec \alpha } \right)$ in equation (1), we get
${\text{LHS}} = \dfrac{{\tan \alpha }}{{{{\left( {\sec \alpha } \right)}^2}}} = \dfrac{{\left( {\sin \alpha } \right)\left( {\sec \alpha } \right)}}{{{{\left( {\sec \alpha } \right)}^2}}} = \dfrac{{\sin \alpha }}{{\sec \alpha }} = {\text{ RHS}}$
Clearly, from the above equation we can say that the LHS of the equation which needs to be proved is equal to the RHS of that equation. Hence, that equation holds true for any value of angle $\alpha $.
Note: In this problem, we are also given the value of $\sec \alpha $ which is not used in order to verify the equation which is asked for. However, another approach is we can put the given value of $\sec \alpha = \dfrac{5}{4}$ and with the help of this value we will find the values of $\tan \alpha $ and $\sin \alpha $then substitute these values in the LHS and RHS of the equation which needs to be proved and from there can verify the equation.
Complete step-by-step answer:
Given, $\sec \alpha = \dfrac{5}{4}$
To prove: $\dfrac{{\tan \alpha }}{{1 + {{\left( {\tan \alpha } \right)}^2}}} = \dfrac{{\sin \alpha }}{{\sec \alpha }}$
Taking LHS of the above equation which needs to be proved
i.e., ${\text{LHS}} = \dfrac{{\tan \alpha }}{{1 + {{\left( {\tan \alpha } \right)}^2}}}$
Using the identity $1 + {\left( {\tan \alpha } \right)^2} = {\left( {\sec \alpha } \right)^2}$ , we get
${\text{LHS}} = \dfrac{{\tan \alpha }}{{{{\left( {\sec \alpha } \right)}^2}}}{\text{ }} \to {\text{(1)}}$
As we know that $\tan \alpha = \dfrac{{\sin \alpha }}{{\cos \alpha }}$ and $\sec \alpha = \dfrac{1}{{\cos \alpha }}$ so we can write that $\tan \alpha = \left( {\sin \alpha } \right)\left( {\sec \alpha } \right)$
Now put $\tan \alpha = \left( {\sin \alpha } \right)\left( {\sec \alpha } \right)$ in equation (1), we get
${\text{LHS}} = \dfrac{{\tan \alpha }}{{{{\left( {\sec \alpha } \right)}^2}}} = \dfrac{{\left( {\sin \alpha } \right)\left( {\sec \alpha } \right)}}{{{{\left( {\sec \alpha } \right)}^2}}} = \dfrac{{\sin \alpha }}{{\sec \alpha }} = {\text{ RHS}}$
Clearly, from the above equation we can say that the LHS of the equation which needs to be proved is equal to the RHS of that equation. Hence, that equation holds true for any value of angle $\alpha $.
Note: In this problem, we are also given the value of $\sec \alpha $ which is not used in order to verify the equation which is asked for. However, another approach is we can put the given value of $\sec \alpha = \dfrac{5}{4}$ and with the help of this value we will find the values of $\tan \alpha $ and $\sin \alpha $then substitute these values in the LHS and RHS of the equation which needs to be proved and from there can verify the equation.
Recently Updated Pages
Master Class 10 General Knowledge: Engaging Questions & Answers for Success

Master Class 10 Computer Science: Engaging Questions & Answers for Success

Master Class 10 Science: Engaging Questions & Answers for Success

Master Class 10 Social Science: Engaging Questions & Answers for Success

Master Class 10 Maths: Engaging Questions & Answers for Success

Master Class 10 English: Engaging Questions & Answers for Success

Trending doubts
Truly whole mankind is one was declared by the Kannada class 10 social science CBSE

Explain the three major features of the shiwaliks class 10 social science CBSE

Distinguish between the reserved forests and protected class 10 biology CBSE

What are the public facilities provided by the government? Also explain each facility

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Question An example of homologous organs is a Our arm class 10 biology CBSE
