Answer

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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.

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