If a\ tan{\theta}=b, the value of

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Q.
If $a t a n θ = b$ the value of $\frac{b \sin θ - a \cos θ}{b \sin θ + a \cos θ}$ is

1 $\frac{b - a}{b^{2} + a^{2}}$

2 $\frac{b + a}{b^{2} + a^{2}}$

3 $\frac{b^{2} + a^{2}}{b^{2} - a^{2}}$

4 $\frac{b^{2} - a^{2}}{b^{2} + a^{2}}$

Ans.
(4)

$W e h a v e, t a n θ = \frac{b}{a} G i v e n e x p r e s s i o n i s \frac{b \sin θ - a \cos θ}{b \sin θ + a \cos θ} T a k i n g \cos θ c o m m o n f r o m t h e e x p r e s s i o n, i t g i v e s \frac{b \tan θ - a}{b \tan θ + a} \dots (i) S u b s t i t u t i n g t a n θ = \frac{b}{a} i n e q u a t i o n (i) g i v e s \frac{\frac{b^{2}}{a} - a}{\frac{b^{2}}{a} + a} = \frac{b^{2} - a^{2}}{b^{2} + a^{2}}$

Solution

Q.
If $a t a n θ = b$ the value of $\frac{b \sin θ - a \cos θ}{b \sin θ + a \cos θ}$ is

1 $\frac{b - a}{b^{2} + a^{2}}$

2 $\frac{b + a}{b^{2} + a^{2}}$

3 $\frac{b^{2} + a^{2}}{b^{2} - a^{2}}$

4 $\frac{b^{2} - a^{2}}{b^{2} + a^{2}}$

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Solution

Q. If a tanθ=b the value of bsinθ-acosθbsinθ+acosθ is

1 b-ab2+a2

2 b+ab2+a2

3 b2+a2b2-a2

4 b2-a2b2+a2

Ans. (4) We have, tanθ=baGiven expression is bsinθ-acosθbsinθ+acosθTaking cosθ common from the expression,it gives btanθ-abtanθ+a…(i)Substituting tanθ=bainequation (i) givesb2a-ab2a+a=b2-a2b2+a2

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Q.
If $a t a n θ = b$ the value of $\frac{b \sin θ - a \cos θ}{b \sin θ + a \cos θ}$ is

1 $\frac{b - a}{b^{2} + a^{2}}$

2 $\frac{b + a}{b^{2} + a^{2}}$

3 $\frac{b^{2} + a^{2}}{b^{2} - a^{2}}$

4 $\frac{b^{2} - a^{2}}{b^{2} + a^{2}}$