Q.1 $Prove that \frac{sin 𝜃-cos 𝜃}{sin 𝜃+cos 𝜃}+\frac{sin 𝜃+cos 𝜃}{sin 𝜃-cos 𝜃}=\frac{2}{2sin\mathrm{²} 𝜃-1}$

 

(OR)

 

Q.2 Prove that the parallelogram circumscribing a circle is rhombus.

Given that, $$LHS=\frac{sin 𝜃-cos 𝜃}{sin 𝜃+cos 𝜃}+\frac{sin 𝜃+cos 𝜃}{sin 𝜃-cos 𝜃}$$

Rationalising the denominator we get,  

 $$=\frac{sin 𝜃-cos 𝜃}{sin 𝜃+cos 𝜃}\left(\frac{sin 𝜃-cos 𝜃}{sin 𝜃-cos 𝜃}\right)+\frac{sin 𝜃+cos 𝜃}{sin 𝜃-cos 𝜃}\left(\frac{sin 𝜃+cos 𝜃}{sin 𝜃+cos 𝜃}\right)$$
 $$=\frac{sin\mathrm{²} 𝜃+cos\mathrm{²} 𝜃 - 2sin 𝜃 cos 𝜃}{sin\mathrm{²} 𝜃-\left(1-sin\mathrm{²} 𝜃\right)}+\frac{sin\mathrm{²} 𝜃+cos\mathrm{²} 𝜃 +2sin 𝜃 cos 𝜃}{sin\mathrm{²} 𝜃-\left(1-sin\mathrm{²} 𝜃\right)}$$

 $$=\frac{1 - 2sin 𝜃 cos 𝜃}{sin\mathrm{²} 𝜃-1+sin\mathrm{²} 𝜃}+\frac{1+2sin 𝜃 cos 𝜃}{sin\mathrm{²} 𝜃-1+sin\mathrm{²} 𝜃}$$ 

 $$=\frac{1 - 2sin 𝜃 cos 𝜃}{2sin\mathrm{²} 𝜃-1}+\frac{1+2sin 𝜃 cos 𝜃}{2sin\mathrm{²} 𝜃-1}$$

 $$=\frac{1-2sin 𝜃 cos 𝜃+1+2sin 𝜃 cos 𝜃}{2sin\mathrm{²} 𝜃-1}$$ 

$=\frac{2}{2sin\mathrm{²} 𝜃-1}$ 

$$= RHS$$

 $$\Rightarrow \frac{sin 𝜃-cos 𝜃}{sin 𝜃+cos 𝜃}+\frac{sin 𝜃+cos 𝜃}{sin 𝜃-cos 𝜃}=\frac{2}{2sin\mathrm{²} 𝜃-1}$$

Hence proved.  

(OR)

Given that,  

Parallelogram ABCD  is circumscribing the circle and its sides are touching the circle at P, Q, R and S.

AP and AS  are tangent to the circle from the external point A.

BP and BQ  are tangent to the circle from the external point B.

CQ and CR are tangents to the circle from the external point C.

DR  and DS  are tangents to the circle from the external point D.

We know that tangents drawn to a circle from an external point are equal. 

$AP=AS.....\left(i\right)$

$BP=BQ......\left(ii\right)$

$CQ=CR.....\left(iii\right)$

$DR=DS.....\left(iv\right)$

Adding

 (1), (2), (3) and (4) we get,  

AP+BP+CR+DR=AS+BQ+CQ+DS

⇒(AP+BP)+(CR+DR)=(AS+DS)+(BQ+CQ)

⇒AB+CD=AD+BC

Since$$ABCD is a parallelogram then AB=CD and BC=AD$$

⇒AB+AB=BC+BC

⇒2AB=2BC

Thus AB=BC=CD=AD

It is a rhombus 

The parallelogram circumscribing a circle is rhombus. 

Hence proved