If tan - - a-b then cos-sin-cos-sin- - - a a-b-a-b b a-b-a-b c b-a-b-a d b-a-b-a

tan θ =a/ b sinθ/ cosθ = a/ b——(1) LHS = cosθ+sinθ/ cosθ sinθ dividing each term with cosθ = cosθ/ cosθ+ sinθ/ cosθ/cosθ/ cosθ sinθ/ cosθ =sinθ/ cosθ+1/1 s ...

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Standard X

Mathematics

Relation between Trigonometric Ratios

If tan θ = a/...

Question

If tan $θ$ = $\frac{a}{b} t h e n \frac{(c o s θ + s i n θ)}{(c o s θ - s i n θ)}$ = ?

(a) $\frac{a + b}{a - b}$ (b) $\frac{a - b}{a + b}$ (c) $\frac{b + a}{b - a}$ (d) $\frac{b - a}{b + a}$

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Solution

tan θ = $\frac{a}{b}$

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

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

dividing each term with cosθ

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

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

= $\frac{\frac{a}{b} + 1}{1 - \frac{a}{b}}$

= $\frac{(b + a)}{(b - a)}$

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If tan θ = abthen(cos θ+sin θ)(cos θ−sin θ) = ? (a) a+ba−b (b) a−ba+b (c) b+ab−a (d) b−ab+a

tan θ =ab sinθcosθ = ab------(1) LHS = cosθ+sinθcosθ−sinθ dividing each term with cosθ = cosθcosθ+sinθcosθcosθcosθ−sinθcosθ =sinθcosθ+11−sinθcosθ =ab+11−ab =(b+a)(b−a)

If tan $θ$ = $\frac{a}{b} t h e n \frac{(c o s θ + s i n θ)}{(c o s θ - s i n θ)}$ = ?

(a) $\frac{a + b}{a - b}$ (b) $\frac{a - b}{a + b}$ (c) $\frac{b + a}{b - a}$ (d) $\frac{b - a}{b + a}$