Question

If cot $\theta$ + cosec $\theta$ = a , then prove that $\bf{(a^2+1)cos\theta+(a^2+1)sin\theta=(a+1)^2-2$​

Answer 1

ANSWER:

Given:

• cot θ + cosec θ = a

To Prove:

• (a² + 1)cos θ + (a² + 1)sin θ = (a + 1)² - 2

Proof:

We are given that,

$\implies \cot\theta+\csc\theta=a$

We need to prove that,

$\implies (a^2+1)\cos\theta+(a^2+1)\sin\theta=(a+1)^2-2$

Taking LHS,

$\implies (a^2+1)\cos\theta+(a^2+1)\sin\theta$

Now, we take (a² + 1) common,

$\implies (a^2+1)(\cos\theta+\sin\theta)$

Substituting the value of a,

$\implies [(\cot\theta+\csc\theta)^2+1][\cos\theta+\sin\theta]$

As,

$\hookrightarrow (x+y)^2=x^2+y^2+2xy$

So,

$\implies [(\cot\theta)^2+(\csc\theta)^2+2(\cot\theta)(\csc\theta)+1][\cos\theta+\sin\theta]$

$\implies [\cot^2\theta+\csc^2\theta+2\cot\theta\csc\theta+1][\cos\theta+\sin\theta]$

As,

$\hookrightarrow \cot\phi=\dfrac{\cos\phi}{\sin\phi}\:\:\&\:\:\csc\phi=\dfrac{1}{\sin\phi}$

So,

$\implies [\cot^2\theta+\csc^2\theta+2\cot\theta\csc\theta+1][\cos\theta+\sin\theta]$

$\implies \left[\left(\dfrac{\cos\theta}{\sin\theta}\right)^2+\left(\dfrac{1}{\sin\theta}\right)^2+2\left(\dfrac{\cos\theta}{\sin\theta}\right)\left(\dfrac{1}{\sin\theta}\right)+1\right]\bigg[\cos\theta+\sin\theta\bigg]$

$\implies \left[\dfrac{\cos^2\theta}{\sin^2\theta}+\dfrac{1}{\sin^2\theta}+\dfrac{2\cos\theta}{\sin^2\theta}+1\right]\bigg[\cos\theta+\sin\theta\bigg]$

Taking LCM,

$\implies \left(\dfrac{\cos^2\theta+1+2\cos\theta+\sin^2\theta}{\sin^2\theta}\right)\bigg(\cos\theta+\sin\theta\bigg)$

Rearranging,

$\implies \left(\dfrac{\cos^2\theta+\sin^2\theta+1+2\cos\theta}{\sin^2\theta}\right)\bigg(\cos\theta+\sin\theta\bigg)$

As,

$\hookrightarrow\cos^2\phi+\sin^2\phi=1$

So,

$\implies \left(\dfrac{(\cos^2\theta+\sin^2\theta)+1+2\cos\theta}{\sin^2\theta}\right)\bigg(\cos\theta+\sin\theta\bigg)$

$\implies \left(\dfrac{1+1+2\cos\theta}{\sin^2\theta}\right)\bigg(\cos\theta+\sin\theta\bigg)$

$\implies \left(\dfrac{2+2\cos\theta}{\sin^2\theta}\right)\bigg(\cos\theta+\sin\theta\bigg)$

$\implies \dfrac{(2+2\cos\theta)(\cos\theta+\sin\theta)}{\sin^2\theta}$

Opening the brackets,

$\implies \dfrac{2\cos\theta+2\sin\theta+2\cos^2\theta+2\cos\theta\sin\theta}{\sin^2\theta}$

As,

$\hookrightarrow\cos^2\phi=1-\sin^2\phi$

So,

$\implies \dfrac{2\cos\theta+2\sin\theta+2\cos^2\theta+2\cos\theta\sin\theta}{\sin^2\theta}$

$\implies \dfrac{2\cos\theta+2\sin\theta+2(1-\sin^2\theta)+2\cos\theta\sin\theta}{\sin^2\theta}$

$\implies \dfrac{2\cos\theta+2\sin\theta+2-2\sin^2\theta+2\cos\theta\sin\theta}{\sin^2\theta}$

We can rewrite it as,

$\implies \dfrac{2+2\cos\theta\sin\theta+2\cos\theta+2\sin\theta-2\sin^2\theta}{\sin^2\theta}$

$\implies \dfrac{1+1+2\cos\theta\sin\theta+2\cos\theta+2\sin\theta-2\sin^2\theta}{\sin^2\theta}$

As,

$\hookrightarrow1=\cos^2\phi+\sin^2\phi$

So,

$\implies \dfrac{1+(1)+2\cos\theta\sin\theta+2\cos\theta+2\sin\theta-2\sin^2\theta}{\sin^2\theta}$

$\implies \dfrac{1+\sin^2\theta+\cos^2\theta+2\cos\theta\sin\theta+2\cos\theta+2\sin\theta-2\sin^2\theta}{\sin^2\theta}$

$\implies \dfrac{\sin^2\theta+\cos^2\theta+2\cos\theta\sin\theta+1+2\cos\theta+2\sin\theta-2\sin^2\theta}{\sin^2\theta}$

As,

$\hookrightarrow x^2+y^2+2xy=(x+y)^2$

So,

$\implies \dfrac{(\sin^2\theta+\cos^2\theta+2\cos\theta\sin\theta)+1+2\cos\theta+2\sin\theta-2\sin^2\theta}{\sin^2\theta}$

$\implies \dfrac{(\sin\theta+\cos\theta)^2+1+2\cos\theta+2\sin\theta-2\sin^2\theta}{\sin^2\theta}$

$\implies \dfrac{(\sin\theta+\cos\theta)^2+1+2(\cos\theta+\sin\theta)-2\sin^2\theta}{\sin^2\theta}$

$\implies \dfrac{(\sin\theta+\cos\theta)^2+1+2(\sin\theta+\cos\theta)-2\sin^2\theta}{\sin^2\theta}$

We can rewrite it as,

$\implies \dfrac{(\sin\theta+\cos\theta)^2+1^2+2(\sin\theta+\cos\theta)(1)-2\sin^2\theta}{\sin^2\theta}$

As,

$\hookrightarrow x^2+y^2+2xy=(x+y)^2$

So,

$\implies \dfrac{[(\sin\theta+\cos\theta)^2+1^2+2(\sin\theta+\cos\theta)(1)]-2\sin^2\theta}{\sin^2\theta}$

$\implies \dfrac{(\sin\theta+\cos\theta+1)^2-2\sin^2\theta}{\sin^2\theta}$

So,

$\implies \dfrac{(\sin\theta+\cos\theta+1)^2}{\sin^2\theta}-\dfrac{2\sin^2\theta}{\sin^2\theta}$

On simplifying,

$\implies \dfrac{(\sin\theta+\cos\theta+1)^2}{\sin^2\theta}-2$

$\implies \left(\dfrac{\sin\theta+\cos\theta+1}{\sin\theta}\right)^2-2$

So,

$\implies \left(\dfrac{\sin\theta}{\sin\theta}+\dfrac{\cos\theta}{\sin\theta}+\dfrac{1}{\sin\theta}\right)^2-2$

As,

$\hookrightarrow \dfrac{\cos\phi}{\sin\phi}=\cot\phi\:\:\&\:\:\dfrac{1}{\sin\phi}=\csc\phi$

So,

$\implies \left(1+\dfrac{\cos\theta}{\sin\theta}+\dfrac{1}{\sin\theta}\right)^2-2$

$\implies (1+\cot\theta+\csc\theta)^2-2$

$\implies (\cot\theta+\csc\theta+1)^2-2$

But, we are given that,

$\implies \cot\theta+\csc\theta=a$

So,

$\implies (\cot\theta+\csc\theta+1)^2-2$

$\implies (a+1)^2-2$

But,

$\implies RHS=(a+1)^2-2$

As, LHS=RHS,

HENCE PROVED!!!