Question

If p=2sintheta/1+costheta+sintheta and q=costheta/1+sintheta then -
a) pq=1 b) q/p=1
c) q-p=1 d) q+p=1

Answer 1

GIVEN :

If $p=\frac{2sin\theta}{1+cos\theta+sin\theta}$ and $q=\frac{cos\theta}{1+sin\theta}$

a) pq=1 b) $\frac{q}{p}=1$  c) q-p=1  and d) q+p=1

TO FIND :

The correct option satisfying the given equations

SOLUTION :

Given that  $p=\frac{2sin\theta}{1+cos\theta+sin\theta}$ and $q=\frac{cos\theta}{1+sin\theta}$

Now rationalise the denominator's of the given value

$p=\frac{2sin\theta}{1+cos\theta+sin\theta}$

$=\frac{2sin\theta}{1+(cos\theta+sin\theta)}$

Multiply and dividing by the denominator's conjugate we get

$=\frac{2sin\theta}{1+(cos\theta+sin\theta)}\times \frac{1-(cos\theta+sin\theta)}{1-(cos\theta+sin\theta)}$

By using the Algebraic identity :

$(a+b)(a-b)=a^2-b^2$

$=\frac{2sin\theta(1-(cos\theta+sin\theta)}{1^2-(cos\theta+sin\theta)^2}$

By using the Algebraic identity :

$(a+b)^2=a^2+2ab+b^2$

$=\frac{2sin\theta(1)-(2sin\theta)(cos\theta)-(2sin\theta)sin\theta}{1-(cos^2\theta+sin^2\theta+2sin\theta cos\theta)}$

By using the Trignometric identity :

$cos^2x+sin^2x=1$

$=\frac{2sin\theta-2sin\theta cos\theta-2sin\theta^2}{1-(1+2sin\theta cos\theta)}$

$=\frac{2sin\theta-2sin\theta cos\theta-2sin\theta^2}{1-1-2sin\theta cos\theta}$

$=\frac{2sin\theta-2sin\theta cos\theta-2sin\theta^2}{-2sin\theta cos\theta}$

$=\frac{2sin\theta(1-(cos\theta+sin\theta))}{-2sin\theta cos\theta}$

$=\frac{1-(cos\theta+sin\theta)}{-cos\theta}$

⇒ $p=\frac{cos\theta+sin\theta-1}{cos\theta}$

Now $q=\frac{cos\theta}{1+sin\theta}$

Now rationalise the denominator's of the given value

$q=\frac{cos\theta}{1+sin\theta}$

Multiply and dividing by the denominator's conjugate we get

$=\frac{cos\theta}{1+sin\theta}\times \frac{1-sin\theta}{1-sin\theta}$

By using the Algebraic identity :

$(a+b)(a-b)=a^2-b^2$

$=\frac{cos\theta(1-sin\theta)}{1^2-sin^2\theta}$

$=\frac{cos\theta(1-sin\theta)}{1-sin^2\theta}$

By using the Trignometric  identity :

$cos^2x=1-sin^2x$

$=\frac{cos\theta(1-sin\theta)}{cos^2\theta}$

⇒ $q=\frac{1-sin\theta}{cos\theta}$

Now verify the option d) q+p=1

Putting the values in the equation q+p=1 we get

LHS $q+p=\frac{1-sin\theta}{cos\theta}+\frac{cos\theta+sin\theta-1}{cos\theta}$

$=\frac{1-sin\theta+cos\theta+sin\theta-1}{cos\theta}$

$=\frac{cos\theta}{cos\theta}$

=1 = RHS

LHS=RHS

∴ q+p=1 is verified and is true

∴ option d) q+p=1 is correct.