Question

 If Tan ⊖ = a∕b, then (a sin⊖ + b cos⊖) / (a sin⊖ - b cos⊖) = ? ​

Answer 1

$\texttt{\textsf{\large{\underline{Solution}:}}}$

Given:

$\tt \implies \tan \theta = \dfrac{a}{b}$

We have to evaluate the given expression.

Given Expression,

$\tt = \dfrac{a \sin \theta + b \cos \theta}{a \sin \theta - b \cos \theta}$

Dividing both numerator and denominator by cos θ, we get,

$\tt = \dfrac{ \dfrac{a \sin \theta + b \cos \theta}{ \cos \theta }}{ \dfrac{a \sin \theta - b \cos \theta}{ \cos \theta}}$

$\tt = \dfrac{a \dfrac{\sin \theta}{\cos \theta} + b \dfrac{\cos \theta}{ \cos \theta}}{a \dfrac{\sin \theta}{\cos \theta} - b \dfrac{\cos \theta}{ \cos \theta}}$

As we know that,

$\tt \implies \tan \theta = \dfrac{\sin \theta}{ \cos \theta}$

Therefore, we get,

$\tt = \dfrac{a \tan \theta + b }{a \tan \theta - b }$

Substituting the value of tan θ, we get,

$\tt = \dfrac{a \times \dfrac{a}{b} + b }{a \times \dfrac{a}{b} - b }$

$\tt = \dfrac{ \dfrac{ {a}^{2} }{b} + b }{ \dfrac{ {a}^{2} }{b} - b }$

$\tt = \dfrac{ \dfrac{ {a}^{2} + {b}^{2} }{b}}{ \dfrac{ {a}^{2} - {b}^{2} }{b}}$

Cancelling out b, we get,

$\tt = \dfrac{ {a}^{2} + {b}^{2} }{{a}^{2} - {b}^{2} }$

Which is our required answer.

$\texttt{\textsf{\large{\underline{Learn More Formulae}:}}}$

1. Relationship between sides.

• sin(x) = Height/Hypotenuse.
• cos(x) = Base/Hypotenuse.
• tan(x) = Height/Base.
• cot(x) = Base/Height.
• sec(x) = Hypotenuse/Base.
• cosec(x) = Hypotenuse/Height.

2. Square Formulae.

• sin²x + cos²x = 1.
• cosec²x - cot²x = 1.
• sec²x - tan²x = 1

3. Reciprocal Relationship.

• sin(x) = 1/cosec(x).
• cos(x) = 1/sec(x).
• tan(x) = 1/cot(x).

4. Cofunction Identities.

• sin(90° - x) = cos(x) and vice versa.
• cosec(90° - x) = sec(x) and vice versa.
• tan(90° - x) = cot(x) and vice versa.

5. Quotient Relations.

• tan(x) = sin(x)/cos(x)
• cot(x) = 1/tan(x) = cos(x)/sin(x)

Answer 2

Answer:

Given,

$\sf tan \theta = \frac{a}{b}$

We know that,

$\sf \: tan\theta = \frac{sin\theta}{cos\theta }$

Now,

$\orange{\sf \: \frac{a \: sin\theta + b \: cos\theta}{a \: sin\theta - b \: cos\theta}} \\ \\ \pink{\sf = \frac{ \frac{a \: sin\theta}{cos\theta} + \frac{b \: cos\theta}{cos\theta} }{\frac{a \: sin\theta}{cos\theta} - \frac{b \: cos\theta}{cos\theta}}} \\ \\ \sf = \frac{a \: tan\theta + b}{a \: tan\theta - b} \\ \\ \sf = \frac{a \times \frac{a}{b} + b}{a \times \frac{a}{b} - b } \\ \\ \sf = \frac{ \frac{ {a}^{2} + {b}^{2} }{\cancel{b}} }{ \frac{ {a}^{2} - {b}^{2} }{\cancel{b}} } \\ \\ \sf = \frac{ {a}^{2} + {b}^{2} }{ {a}^{2} - {b}^{2} }$

So,

$\large \green{ \boxed{ \sf \: \frac{a \: sin\theta + b \: cos\theta}{a \: sin\theta - b \: cos\theta} = \frac{ {a}^{2} + {b}^{2} }{ {a}^{2} - {b}^{2} } }}$