Question

please solve the question in the attachment
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Answer 1

To Find:

$( \frac{1 + \sin(θ) }{1 + \cos(θ) } )( \frac{1 - \sin(θ) }{1 - \cos(θ) })$

Using,

$(x + y)(x - y) = {x}^{2} - {y}^{2}$

The Expression Simplifies to

$\frac{1 - \sin^{2} (θ) }{1 - \cos ^{2} (θ) }$

Now, We know that

$\sin^{2} (θ) + \cos^{2} (θ) = 1$

Therefore, the Expression Simplifies to

$= \frac{ \cos^{2} (θ)}{ \sin^{2} (θ)} = \cot^{2} (θ)$

Taking C as θ,

$\cot^{2} (θ) = (\frac{7k}{8k}) ^{2} \\ \cot^{2} (θ) = \frac{49}{64}$

Taking A as θ,

$\cot^{2} (θ) = (\frac{8k}{7k}) ^{2} \\ \cot^{2} (θ) = \frac{64}{49}$

Hope It Helps

Answer 2

According to the statement which you have to find value of , use identity (a+b)(a-b) = a^2 - b^2

Now in the numerator you have a= 1 , b= sin(theta)

Therefore you will have 1 - sin^2(theta) in the numerator

Similarly you will have 1- cos^2 (theta) in the denominator


Now there is an trigonometric identity according to which sin^2(theta) + cos^2(theta) =1

Therefore you will have cos^2(theta) in numerator [by substituting the values from above equation]
And sin^2(theta) in denominator [ by substituting the values from the above equation]

So finally you have to find the values of [cos^2(theta)]/[sin^2(theta)]

Which is equal to cot^2theta , (by simple trigonometric ratios )

Therefore now cot theta = alternate / opposite

But in the diagram which you have provided , it doesn’t contain any angle named as theta in the triangle

Now hope it may helped you , please mark this answer as brainiest answer if you understood, and plzz like it (iss gareeb ka kuch bhala karo yrr)