Question

how to solve above question​

Answer 1

★ Concept :-

Here the concept of Trigonometry has been used. We see that we are given one initial expression and we need to find a final one. Firstly we can derive the specific values from the initial equation. Then we can take the final equation and there simplify it to find our answer. And then the values which we got, we can apply it and hence get the answer.

Let's do it !!

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★ Formula Used :-

$\;\boxed{\sf{\pink{\sin^{2}\theta\;+\;\cos^{2}\:theta\;=\;1}}}$

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★ Solution :-

Given,

$\;\bf{\odot\;\;\red{\dfrac{\sin^{2}\alpha}{x}\;=\;\dfrac{\cos^{2}\alpha}{y}}}$

Now here let these both be equal to a constant k. So,

$\;\bf{\odot\;\;\orange{\dfrac{\sin^{2}\alpha}{x}\;=\;\dfrac{\cos^{2}\alpha}{y}\;=\;k}}$

Now we have to find the value of equation as,

$\;\sf{\Longrightarrow\;\;\green{\dfrac{\sin^{2}\alpha}{x^{2}}\:+\:\dfrac{\cos^{2}\alpha}{y^{2}}\;=\;?}}$

This equation can be written as,

$\;\bf{\Longrightarrow\;\;\bigg(\dfrac{1}{x}\:\times\:\dfrac{\sin^{2}\alpha}{x}\bigg)\:+\:\bigg(\dfrac{1}{y}\:\times\:\dfrac{\cos^{2}\alpha}{y}\bigg)}$

Since we already assumed the quantities as k so by applying that, we get

$\;\bf{\Longrightarrow\;\;\bigg(\dfrac{1}{x}\:\times\:k\bigg)\:+\:\bigg(\dfrac{1}{y}\:\times\:k\bigg)}$

$\;\bf{\Longrightarrow\;\;\dfrac{k}{x}\:+\:\dfrac{k}{y}}$

Now adding these both we get,

$\;\bf{\Longrightarrow\;\;\dfrac{ky\;+\;kx}{xy}}$

Now applying the value of k according to the multiplied terms, we get

$\;\bf{\Longrightarrow\;\;\dfrac{\bigg(\frac{\cos^{2}\alpha}{y}\:\times\:y\bigg)\;+\;\bigg(\frac{\sin^{2}\alpha}{x}\:\times\:x\bigg)}{xy}}$

Cancelling the respective terms, we get

$\;\bf{\Longrightarrow\;\;\dfrac{\cos^{2}\alpha\;+\;\sin^{2}\alpha}{xy}}$

$\;\bf{\Longrightarrow\;\;\dfrac{\sin^{2}\alpha\;+\;\cos^{2}\alpha}{xy}}$

From formula we know that,

$\;\sf{\rightarrow\;\;\sin^{2}\theta\;+\;\cos^{2}\:theta\;=\;1}$

• Here sin²θ = sin² α

• Here cos²θ = cos² α

By applying this, we get

$\;\bf{\Longrightarrow\;\;\blue{\dfrac{1}{xy}}}$

On applying the required equation, we get

$\;\bf{\Longrightarrow\;\;\blue{\dfrac{\sin^{2}\alpha}{x^{2}}\:+\:\dfrac{\cos^{2}\alpha}{y^{2}}\;=\;\dfrac{1}{xy}}}$

This is the required answer

So option b.) 1/xy is correct.

$\;\underline{\boxed{\tt{Required\;\:answer\;=\;\bf{\purple{\dfrac{1}{xy}}}}}}$

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★ More to know :-

• cosec² θ = 1 + cot² θ

• sec² θ = 1 + tan² θ

• sin θ = cos(90° - θ)

• tan θ = cot(90° - θ)

• cosec θ = sec(90° - θ)

Answer 2

Answer:

Option (b) 1/xy is correct

Step-by-step explanation:

★Given:-

sin²α / x= cos ²α /y

★To find : -

The value of

sin²α/x²+cos²α/y²

★Solution:

Let sin²α / x= cos ²α/y=k

Then sin²α=kx and cos²α=ky

⇒ kx+ky=sin²α+cos²α=1

⇒ k(x+y)=1-----------------------(1)

Now

sin²α/x²+cos²α/y²

=kx/x²+ky/y²

=k/x +k/y

=k(x+y)/xy

From(1) putting the value of k(x+y)

sin²α/x²+cos²α/y²=k(x+y)/xy

=1/xy

★Answer:-

Option (b) 1/xy is correct

★Formulas used:

1. sin²α+cos²α=1
2. If a/b=c/d=k then a=bk and c=dk

★★ More to know

why sin²α+cos²α=1 ?

We know that sinα=height /hypotenuse

and cosα=base/hypotenuse

Thus sin²α+cos²α

=(height /hypotenuse)²+(base /hypotenuse)²

=(height²+base²)/hypotenuse²

=hypotenuse²/hypotenuse² ( as Height²+base²=hypotenuse²)

=1

Other related formulas

• sec²α=1+tan²α

• cosec²α=1+cot²α