Question

Prove the following:
( I ) \[\sec \theta (1 - \sin \theta )(\sec \theta + \tan \theta ) = 1\]
( ii ) \[{\cot ^2}\theta - {\tan ^2}\theta = \cos e{c^2}\theta - {\sec ^2}\theta \]

Answer 1

Hint: Here the question is related to the trigonometry topic. We have to prove the following trigonometric function. While solving we use the trigonometric identities and hence we determine the solution or we prove the given trigonometric functions.

Complete step by step answer:
The question is related to trigonometry and it includes the trigonometry ratios. The trigonometry ratios are sine, cosine, tan, cosec, sec, and cot.
Now consider the first sub question
( I ) \[\sec \theta (1 - \sin \theta )(\sec \theta + \tan \theta ) = 1\]
Consider the LHS
\[\sec \theta (1 - \sin \theta )(\sec \theta + \tan \theta )\]
First we multiply the \[\sec \theta \] to the \[(1 - \sin \theta )\] we get
\[ \Rightarrow \sec \theta - \sec \theta \sin \theta (\sec \theta + \tan \theta )\]
We know that the \[\sec \theta = \dfrac{1}{{\cos \theta }}\] so the above equation is written as
\[ \Rightarrow \sec \theta - \dfrac{{\sin \theta }}{{\cos \theta }}(\sec \theta + \tan \theta )\]
We know that the \[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\] so the above equation is written as
\[ \Rightarrow (\sec \theta - \tan \theta )(\sec \theta + \tan \theta )\]
The equation is in the form of \[(a - b)(a + b)\], we have the algebraic formula \[(a - b)(a + b) = {a^2} - {b^2}\], so the equation can be written as
\[ \Rightarrow ({\sec ^2}\theta - {\tan ^2}\theta )\]
We know that the trigonometric identities \[{\sec ^2}\theta = 1 + {\tan ^2}\theta \], the equation is written as
\[ \Rightarrow (1 + {\tan ^2}\theta - {\tan ^2}\theta )\]
On cancelling the \[{\tan ^2}\theta \] we get
\[ \Rightarrow 1\]
\[ = RHS\]
Hence we proved LHS = RHS
Now we prove the second sub question.
Now consider the second sub question
\[{\cot ^2}\theta - {\tan ^2}\theta = \cos e{c^2}\theta - {\sec ^2}\theta \]
Consider the LHS we have
\[ \Rightarrow {\cot ^2}\theta - {\tan ^2}\theta \]
We know the trigonometric identities \[1 + {\tan ^2}\theta = {\sec ^2}\theta \] and \[1 + {\cot ^2}\theta = \cos e{c^2}\theta \], by using these trigonometric identities the above equation is written as
\[ \Rightarrow \cos e{c^2}\theta - 1 - ({\sec ^2}\theta - 1)\]
on simplifying we have
\[ \Rightarrow \cos e{c^2}\theta - 1 - {\sec ^2}\theta + 1\]
The +1 and -1 will gets cancels so we have
\[ \Rightarrow \cos e{c^2}\theta - {\sec ^2}\theta \]
\[ \Rightarrow RHS\]
Hence we proved LHS = RHS

Note: The question involves the trigonometric functions and we have to prove the trigonometric function. When we simplify the trigonometric functions and which will be equal to the RHS then the function is proved. While simplifying the trigonometric functions we must know about the trigonometric ratios and the trigonometric identities.