Question

The value of ${{\tan }^{2}}A-{{\sec }^{2}}A$ is

Answer 1

Hint: For answering this question we need to find the relation between ${{\tan }^{2}}A$ and ${{\sec }^{2}}A$ and define the trigonometric identity using the definitions of the trigonometric ratios. From the definition of $\tan A$ the value is given as $\tan A=\left( \dfrac{\text{opposite side of A}}{\text{adjacent side of A}} \right)$ and for $\sec A$ the value is given as $\sec A=\left( \dfrac{\text{hypotenuse of A}}{\text{adjacent side of A}} \right)$ .

Complete step by step answer:
From the basic definitions of the trigonometric ratios we know that the value of $\tan A$ and $\sec A$ is given as $\tan A=\left( \dfrac{\text{opposite side of A}}{\text{adjacent side of A}} \right)$ and $\sec A=\left( \dfrac{\text{hypotenuse of A}}{\text{adjacent side of A}} \right)$ respectively.
From the definition of Pythagoras theorem, we can say that in a right angle triangle, the square of the side opposite to the right angle that is hypotenuse is equal to the sum of the squares of the others two sides that is mathematically given as,
${{\left( \text{opposite side of A} \right)}^{2}}\text{+}{{\left( \text{Adjacent side of A} \right)}^{2}}\text{=}{{\left( \text{hypotenuse of A} \right)}^{2}}$.
For further simplifying the expression given in the question we will substitute the respective values and use the Pythagoras theorem to solve the expression further.
After substituting the values we will have,
${{\tan }^{2}}A-{{\sec }^{2}}A={{\left( \dfrac{\text{opposite side of A}}{\text{adjacent side of A}} \right)}^{2}}-{{\left( \dfrac{\text{hypotenuse of A}}{\text{adjacent side of A}} \right)}^{2}}$ .
After simplifying them we will have,
${{\tan }^{2}}A-{{\sec }^{2}}A=\dfrac{{{\left( \text{opposite side of A} \right)}^{2}}-{{\left( \text{hypotenuse of A} \right)}^{2}}}{{{\left( \text{adjacent side of A} \right)}^{2}}}$ .
From the Pythagoras theorem ${{\left( \text{opposite side of A} \right)}^{2}}\text{+}{{\left( \text{Adjacent side of A} \right)}^{2}}\text{=}{{\left( \text{hypotenuse of A} \right)}^{2}}$ we will have
${{\left( \text{opposite side of A} \right)}^{2}}-{{\left( \text{hypotenuse of A} \right)}^{2}}\text{=}-{{\left( \text{adjacent side of A} \right)}^{2}}$ .
After substituting this value in the expression we have it will be ${{\tan }^{2}}A-{{\sec }^{2}}A=\dfrac{-{{\left( \text{adjacent side of A} \right)}^{2}}}{{{\left( \text{adjacent side of A} \right)}^{2}}}$
After simplifying it will be reduced to ${{\tan }^{2}}A-{{\sec }^{2}}A=-1$ .

We can conclude that the value of ${{\tan }^{2}}A-{{\sec }^{2}}A$ is -1.

Note: While answering questions of this type it would be efficient if we remember all the three trigonometric identities derived in a similar way. Those identities are stated as follows:
${{\sin }^{2}}A+{{\cos }^{2}}A=1$ and ${{\sec }^{2}}A=1+{{\tan }^{2}}A$ and ${{\csc }^{2}}A=1+{{\cot }^{2}}A$ .