If tan - -a-b-prove that a-sin-b-cos-a-sin- -b-cos-a2-b2-a2-b2

Tan θ = a/b sinθ/cos θ = a/b——(1) LHS = a sinθ b cos θ/a sinθ + b cos θ = asinθ/cosθ bcosθ/cosθ/asinθ/cosθ + bcosθ/cosθ dividing each term with cosθ =as ...

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Standard X

Mathematics

Relation between Trigonometric Ratios

If tan θ =a/b...

Question

If tan $θ = \frac{a}{b}$ ,prove that $\frac{a s i n θ - b c o s θ}{a s i n θ + b c o s θ} = \frac{a^{2} - b^{2}}{a^{2} + b^{2}}$

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Solution

Tan θ = $\frac{a}{b}$

$\frac{s i n θ}{c o s θ}$ = $\frac{a}{b}$ ------(1)

LHS = $\frac{a s i n θ - b c o s θ}{a s i n θ + b c o s θ}$

= $\frac{\frac{a s i n θ}{c o s θ} - \frac{b c o s θ}{c o s θ}}{\frac{a s i n θ}{c o s θ} + \frac{b c o s θ}{c o s θ}}$

dividing each term with cosθ

= $\frac{\frac{a s i n θ}{c o s θ} - \frac{b}{1}}{\frac{a s i n θ}{c o s θ} + \frac{b}{1}}$

= $\frac{a * \frac{a}{b} - \frac{b}{1}}{a * \frac{a}{b} + \frac{b}{1}}$
[from (1)]

= $\frac{a^{2} - b^{2}}{a^{2} + b^{2}}$
=RHS

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If tan θ=ab,prove that a sin θ−b cos θa sin θ+b cos θ=a2−b2a2+b2

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If tan $θ = \frac{a}{b}$ ,prove that $\frac{a s i n θ - b c o s θ}{a s i n θ + b c o s θ} = \frac{a^{2} - b^{2}}{a^{2} + b^{2}}$