If $\cot \theta=\frac{7}{8}$, evaluate:
(i) $\frac{(1+\sin \theta)(1-\sin \theta)}{(1+\cos \theta)(1-\cos \theta)}$
(ii) $\cot ^{2} \theta$

Answer

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Precalculus

NCERT Class 10 - Math

Chapter 8

Introduction to Trigonometry

Section 1

Introduction

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Video Transcript

So in the given question we are told that three times God of an angle a is equal to four. And if this is the case, we are told to check whether the Ratio 1- Stand Square 1 -10 square a divided by one plus 10 square a is equal to cost with a minus sine squared A. So the from this given expression, what we can write us caught a sequel to four by three, right? So we can find the value of 10 F. London's that with using the identity that God is equal to one by 10 8. And you think this weekend right? 10 is equal to three x 4? Right? It is the rest of not enough. But Next we have to find the values of course a and sign a right. So there is an identity that says one place tan square eight is equal to six square eight. So over here we can substitute for Tan Square, we can substitute three x 3 x four square which would be nine x 16. So this is equal to one plus nine by 16 is 25 by 16, which is equal to six square eight. And what we can do from this is we can find the value of 68 as the root of 25x16, which is equal to five by four. Next what we can do is we can use the identity that seek A is equal to one by cost A two. Right? That cause they would then be equal to four x 5. So now that we have found cost is value. We can use the identity. The next identity that is cost square A Plus Science square A Is equal to one. So from this we can find the value of Science square A. As one minus course square A. Where we can replace cost A with four by five. So we will have four by five squared which is 16 by 25 right? So this would be equal to nine by 25. And from this we can write sine is equal to Square root of nine x 25, Which is equal to three by fight. So now We have the values of 10 costs and sign right. We have the values of 10 a. Which is equal to three x 4. We have signed A. Which is three x 5 and caused A. Which is equal to four x 5. So now let's start from the left hand side of the of this relation that is given over here. So when we start from the left hand side we have one minor stand square A plus one plus stand square, right. And this Would be equal to let's substitute for 10 S three x 4. And what we would have is one minus nine x 16, divided by 1-plus 9 x 16, which is equal to 16 minus nine, which is seven Divided by 16, divided by 25 divided by 16. And this would then be equal to seven x 25. Now in the right hand side, what we have is past square A minus sine squared A. So we can replace costs a with four x 5, so four by five squared is 16 by 25 minus. We have signed square A. Where we can replace signed A with three by five, so three by five squared is equal to nine by 25 which is equal to seven by 25 to. So what we have found over here is that the left hand side is indeed equal to the left hand side. The left hand side is equal to the right hand side, which means this relation exists and it is equal. The left hand side of this relation is equal to the right hand side. So I hope you understood the method. Thank you.

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