Question

pls.. find and answer with the steps​

Answer 1

$\Large{\underline{\underline{\mathfrak{\tt{\red{Solution}}}}}}$.

$\Large{\underline{\mathfrak{\tt{\orange{Given}}}}}$

• tan θ = x/y

$\Large{\underline{\mathfrak{\tt{\orange{Find}}}}}$

• Value of ( y cos θ + x sin θ)/(x sin θ - y cos θ)

$\Large{\underline{\underline{\mathfrak{\tt{\red{Explanation}}}}}}$.

Using Formula

★ tan θ = Perpendicular/ Base

★ cos θ = Base/ Hypotenuse

★ sin θ =Perpendicular/ Hypotenuse

________________________

So,in Right triangle ∆ ABC

where,

• AB = Perpendicular
• BC = Base
• CA = Hypotenuse

==> tan θ = Perpendicular(AB)/ Base(BC) = x/y

Now, Pythagoras's Theorem

★ Perpendicular² + Base² = Hypotenuse²

Then,

==> Hypotenuse² = x² + y²

==> Hypotenuse(CA) = √(x² + y²)

So,

==> cos θ = Base/ Hypotenuse = y/√(x²+y²)

And,

==> sin θ = Perpendicular/Hypotenuse/ = x/√(x²+y²)

Now, Keep all values

==> ( y cos θ + x sin θ)/(x sin θ - y cos θ)

==> [ y * y/√(x²+y²) + x * x/√(x²+y²)]/[x * x/√(x²+y²) - y * y/√(x²+y²) ]

==> [ (y² + x²)/√(x²+y2) ] / [ (x²-y²)/√(x²+y²)]

==> (x² + y²)/( x ² - y²)

(Ans.)

________________________

Answer 2

$\bigstar$ Answer:

$\bold{tan\theta=\frac{opp.side}{adj.side}=\frac{x}{y}}$

• Now by using pythagoras theorem ,

• (opp.side)²+(adj.side)²=(hypotenuse)²

• (hyp.)² = x² + y²

• hyp. = √(x²+y²)

$\bold{Now,\;sin\theta=\frac{opp.side}{hyp.}=\frac{x}{\sqrt{x^2+y^2}}}$

$\bold{cos\theta=\frac{adj..side}{hyp.}=\frac{y}{\sqrt{x^2+y^2}}}$

• Now we have to find the required value.

$\implies \frac{y.cos\theta+x.\sin\theta}{x.sin\theta-y.cos\theta} \\\\\implies \frac{y.\frac{y}{\sqrt{x^2+y^2}}+x.\frac{x}{\sqrt{x^2+y^2}}}{x.\frac{x}{\sqrt{x^2+y^2}}-y.\frac{y}{\sqrt{x^2+y^2}}}$

$\implies \frac{\frac{y^2}{\sqrt{x^2+y^2}}+\frac{x^2}{\sqrt{x^2+y^2}} }{\frac{x^2}{\sqrt{x^2+y^2}}-\frac{y^2}{\sqrt{x^2+y^2}} } \\\\\implies \frac{\frac{x^2+y^2}{\sqrt{x^2+y^2}} }{\frac{x^2-y^2}{\sqrt{x^2+y^2}} } \\\\\implies \frac{x^2+y^2}{\sqrt{x^2+y^2}}*\frac{\sqrt{x^2+y^2}}{x^2-y^2}\\\\\implies \bold{\frac{x^2+y^2}{x^2-y^2}\;is\;the\;final\;\;answer}$