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.