Question

Evaluate: $\cos e{c^2}(90 - \theta ) - {\tan ^2}\theta $

Answer 1

Hint: First, we see about trigonometry.
Which is the branch of mathematics which is concerned with specific functions of angles with their applications.
There are six types of functions: \[\sin ,\cos ,\tan ,\cos ec,\sec ,\cot \]they are interrelated to each other like $\dfrac{1}{{\sin }} = \cos ec$
These functions are represented in ratio form of the vertical and horizontal lines in the line graph with the x-axis and y-axis.
Formula used: ${[\cos ec(90 - \theta )]^2} = {\sec ^2}\theta $(in ninety-degree the quadrant will be changed), ${\sec ^2}\theta - {\tan ^2}\theta = 1$

Complete step by step answer:
From the given that we have $\cos e{c^2}(90 - \theta ) - {\tan ^2}\theta $
Now we will convert this form as $\cos e{c^2}(90 - \theta ) \Rightarrow {[\cos ec(90 - \theta )]^2}$(taking the square terms common so that we can able apply any trigonometric formula for the changed function in the different quadrant).
Hence, we get the function as $\cos e{c^2}(90 - \theta ) - {\tan ^2}\theta \Rightarrow {[\cos ec(90 - \theta )]^2} - {\tan ^2}\theta $ (taken out the square)
Now acting the ninety degrees minus theta we get the next quadrant in the trigonometry functions like $\sin (90 - \theta ) = \cos \theta $
(The three sides hypotonus, opposite and adjacent sides without the angle of interest is $\sin 90$, so according to the sin function the angle of sin ninety degrees will be equal to the opposite sides ratio of the length of the hypotenuse. Hence it will change to cos in the quadrant one)
With the same method for cosec, $[\cos ec(90 - \theta )] = \dfrac{1}{{\sin (90 - \theta )}} = \dfrac{1}{{\cos \theta }} = \sec \theta $
Now we will get ${[\cos ec(90 - \theta )]^2} - {\tan ^2}\theta \Rightarrow {[\sec \theta ]^2} - {\tan ^2}\theta $
Now giving the square term inside the function and theta value will not get changed, thus we get, ${[\sec \theta ]^2} - {\tan ^2}\theta \Rightarrow {\sec ^2}\theta - {\tan ^2}\theta $
Hence there is a trigonometry formula that sec square minus the tan square yields the number one.
Therefore, we get, ${\sec ^2}\theta - {\tan ^2}\theta = 1$(which is the known result in trigonometry)

Note: We will apply the quadrant formula in the ninety degrees that the values are,
$[\cos ec(90 - \theta )] = \sec \theta $ (Which is the formula that we used in this method to solve further)
${\sin ^2}\theta + {\cos ^2}\theta = 1$(Which is also the well-known formula in the concept of trigonometry and more similar to that we used the formula ${\sec ^2}\theta - {\tan ^2}\theta = 1$)