If tanθ=yx, then xsinθ−ycosθxsinθ+ycosθ is equal to

xsinθ−ycosθxsinθ+ycosθMultiply and divide by cosθ=xcosθsinθ−ycosθcosθxcosθsinθ+ycosθcosθ=x×yx−yx×yx+y=yx2−y2yx2+y2=x2−y2x2+y2

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If ={x}{y}, then {x +y }{x -y } is equal to

Question

If $tan θ = \frac{x}{y},$ then $\frac{x sin θ + y cos θ}{x sin θ - y cos θ}$ is equal to

\frac{x ^{2} + y ^{2}}{x ^{2} - y ^{2}}

\frac{x ^{2} - y ^{2}}{x ^{2} + y ^{2}}

\frac{x}{x ^{2} + y ^{2}}

\frac{y}{x ^{2} + y ^{2}}

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Amelia Discussed

tan θ = \frac{x}{y},

then

\frac{x sin θ + y cos θ}{x sin θ - y cos θ}

is equal to

15 mins ago

Discuss this question LIVE

15 mins ago

\frac{x sin θ + y cos θ}{x sin θ - y cos θ}

Multiply and divide by

cos θ

= \frac{x \frac{sin θ}{cos θ} + y \frac{cos θ}{cos θ}}{x \frac{sin θ}{cos θ} - y \frac{cos θ}{cos θ}}

= \frac{x \times \frac{x}{y} + y}{x \times \frac{x}{y} - y}

= \frac{\frac{x ^{2} + y ^{2}}{y}}{\frac{x ^{2} - y ^{2}}{y}}

= \frac{x ^{2} + y ^{2}}{x ^{2} - y ^{2}}

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Question Text	If $tan θ = \frac{x}{y},$ then $\frac{x sin θ + y cos θ}{x sin θ - y cos θ}$ is equal to
Answer Type	Text solution:1
Upvotes	150