Question

The value of ${{\sec }^{2}}\theta =$
A. $1-{{\cos }^{2}}\theta$
B. $1-{{\tan }^{2}}\theta$
C. $1+{{\tan }^{2}}\theta$
D. $1+{{\cot }^{2}}\theta$

Answer 1

Hint : We use the inverse relations like $\sec \theta =\dfrac{1}{\cos \theta }$. We also have $\dfrac{\sin \theta }{\cos \theta }=\tan \theta$. We use the identity formula of ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ to find the simplified form of ${{\sec }^{2}}$ to find the exact relation with ratio tan.

Complete step-by-step answer :
We know the inverse trigonometric relation $\sec \theta =\dfrac{1}{\cos \theta }$.
Taking square on both sides we get ${{\sec }^{2}}\theta =\dfrac{1}{{{\cos }^{2}}\theta }$.
We now change the value of the numerator using the identity of ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$.
So, we can write ${{\sec }^{2}}\theta =\dfrac{1}{{{\cos }^{2}}\theta }=\dfrac{{{\sin }^{2}}\theta +{{\cos }^{2}}\theta }{{{\cos }^{2}}\theta }$.
Now we need to simplify the division to get the required relation.
We get ${{\sec }^{2}}\theta =\dfrac{{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta }+\dfrac{{{\cos }^{2}}\theta }{{{\cos }^{2}}\theta }=1+\dfrac{{{\sin }^{2}}\theta }{{{\cos }^{2}}\theta }$.
We also have the indices formula of $\dfrac{{{a}^{m}}}{{{b}^{m}}}={{\left( \dfrac{a}{b} \right)}^{m}}$.
Applying the formula, we get ${{\sec }^{2}}\theta =1+{{\left( \dfrac{\sin \theta }{\cos \theta } \right)}^{2}}$.
We know the trigonometric relation of $\dfrac{\sin \theta }{\cos \theta }=\tan \theta$.
The final outcome will be ${{\sec }^{2}}\theta =1+{{\left( \dfrac{\sin \theta }{\cos \theta } \right)}^{2}}=1+{{\tan }^{2}}\theta$.
The correct option is C.
So, the correct answer is “Option C”.

Note : We know that the identity formula of ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$ is called the Pythagoras’ formula. We can also convert to other relations using other trigonometric ratios.