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Question: Prove the following: ( I ) \[\sec \theta (1 - \sin \theta )(\sec \theta + \tan \theta ) = 1\] ( ...

Prove the following:
( I ) sec⁡θ(1−sin⁡θ)(sec⁡θ+tan⁡θ)=1\sec \theta (1 - \sin \theta )(\sec \theta + \tan \theta ) = 1
( ii ) cot⁡2θ−tan⁡2θ=cos⁡ec2θ−sec⁡2θ{\cot ^2}\theta - {\tan ^2}\theta = \cos e{c^2}\theta - {\sec ^2}\theta

Explanation

Solution

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⁡θ(1−sin⁡θ)(sec⁡θ+tan⁡θ)=1\sec \theta (1 - \sin \theta )(\sec \theta + \tan \theta ) = 1
Consider the LHS
sec⁡θ(1−sin⁡θ)(sec⁡θ+tan⁡θ)\sec \theta (1 - \sin \theta )(\sec \theta + \tan \theta )
First we multiply the sec⁡θ\sec \theta to the (1−sin⁡θ)(1 - \sin \theta ) we get
⇒sec⁡θ−sec⁡θsin⁡θ(sec⁡θ+tan⁡θ)\Rightarrow \sec \theta - \sec \theta \sin \theta (\sec \theta + \tan \theta )
We know that the sec⁡θ=1cos⁡θ\sec \theta = \dfrac{1}{{\cos \theta }} so the above equation is written as
⇒sec⁡θ−sin⁡θcos⁡θ(sec⁡θ+tan⁡θ)\Rightarrow \sec \theta - \dfrac{{\sin \theta }}{{\cos \theta }}(\sec \theta + \tan \theta )
We know that the tan⁡θ=sin⁡θcos⁡θ\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }} so the above equation is written as
⇒(sec⁡θ−tan⁡θ)(sec⁡θ+tan⁡θ)\Rightarrow (\sec \theta - \tan \theta )(\sec \theta + \tan \theta )
The equation is in the form of (a−b)(a+b)(a - b)(a + b), we have the algebraic formula (a−b)(a+b)=a2−b2(a - b)(a + b) = {a^2} - {b^2}, so the equation can be written as
⇒(sec⁡2θ−tan⁡2θ)\Rightarrow ({\sec ^2}\theta - {\tan ^2}\theta )
We know that the trigonometric identities sec⁡2θ=1+tan⁡2θ{\sec ^2}\theta = 1 + {\tan ^2}\theta , the equation is written as
⇒(1+tan⁡2θ−tan⁡2θ)\Rightarrow (1 + {\tan ^2}\theta - {\tan ^2}\theta )
On cancelling the tan⁡2θ{\tan ^2}\theta we get
⇒1\Rightarrow 1
=RHS= RHS
Hence we proved LHS = RHS
Now we prove the second sub question.
Now consider the second sub question
cot⁡2θ−tan⁡2θ=cos⁡ec2θ−sec⁡2θ{\cot ^2}\theta - {\tan ^2}\theta = \cos e{c^2}\theta - {\sec ^2}\theta
Consider the LHS we have
⇒cot⁡2θ−tan⁡2θ\Rightarrow {\cot ^2}\theta - {\tan ^2}\theta
We know the trigonometric identities 1+tan⁡2θ=sec⁡2θ1 + {\tan ^2}\theta = {\sec ^2}\theta and 1+cot⁡2θ=cos⁡ec2θ1 + {\cot ^2}\theta = \cos e{c^2}\theta , by using these trigonometric identities the above equation is written as
⇒cos⁡ec2θ−1−(sec⁡2θ−1)\Rightarrow \cos e{c^2}\theta - 1 - ({\sec ^2}\theta - 1)
on simplifying we have
⇒cos⁡ec2θ−1−sec⁡2θ+1\Rightarrow \cos e{c^2}\theta - 1 - {\sec ^2}\theta + 1
The +1 and -1 will gets cancels so we have
⇒cos⁡ec2θ−sec⁡2θ\Rightarrow \cos e{c^2}\theta - {\sec ^2}\theta
⇒RHS\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.