Question

If cot theta =q/p , show that p sin theta - q cos theta /p sin theta +q cos theta =p^2 -q^2/p^2+q^2 ​

Answer 1

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

GIVEN,

$\tt \mapsto \cot(x) = \dfrac{q}{p}$

We have to show that –

$\tt \mapsto \dfrac{p \sin(x) - q \cos(x) }{p \sin(x) + q \cos(x) } = \dfrac{{p}^{2} - {q}^{2}}{ {p}^{2} + {q}^{2} }$

We know that,

$\tt \mapsto \cot(x) = \dfrac{1}{ \tan(x) }$

Also,

$\tt \mapsto \tan(x) = \dfrac{ \sin(x) }{ \cos(x) }$

So,

$\tt \mapsto \cot(x) = \dfrac{ \sin(x)}{ \cos(x) }$

Now,

$\tt = \dfrac{p \sin(x) - q \cos(x) }{p \sin(x) + q \cos(x) }$

Dividing both numerator and denominator by sin(x), we get,

$\tt = \dfrac{\dfrac{p \sin(x) - q \cos(x) }{\sin(x)}}{\dfrac{p \sin(x) + q \cos(x)}{\sin(x)}}$

$\tt = \dfrac{p - \dfrac{q \cos(x) }{\sin(x)}}{p+\dfrac{ q \cos(x)}{\sin(x)}}$

Now, cot(x) = cos(x)/sin(x). So, the fraction becomes,

$\tt = \dfrac{p - q \times \dfrac{q}{p}}{p+q \times\dfrac{ q}{p}}$

$\tt = \dfrac{\dfrac{p^{2} - q^{2}}{p}}{\dfrac{p^{2}+q^{2}}{p}}$

Now, we can cancel out p. We get -

$\tt = \dfrac{p^{2} - q^{2}}{p^{2}+q^{2}}$

Thus,

$\tt \mapsto \dfrac{p \sin(x) - q \cos(x) }{p \sin(x) + q \cos(x) } = \dfrac{{p}^{2} - {q}^{2}}{ {p}^{2} + {q}^{2} }$

Hence, Proved.

$\texttt{\textsf{\large{\underline{Additional Information}:}}}$

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.