Question

If $\sin \left( \theta +\alpha \right)=a$ and $\sin \left( \theta +\beta \right)=b$ , $\left( \left( 0<\alpha ,\beta ,\theta <\dfrac{\pi }{2} \right) \right)$ then $2{{\cos }^{2}}\left( \alpha -\beta \right)-1-4ab\cos \left( \alpha -\beta \right)$ is
(A)$1-{{a}^{2}}-{{b}^{2}}$
(B)$1-2{{a}^{2}}-2{{b}^{2}}$
(C)$2+{{a}^{2}}+{{b}^{2}}$
(D)$2-{{a}^{2}}-{{b}^{2}}$

Answer 1

Hint: For answering this question we will first find the value of $\cos \left( \theta +\alpha \right)$ and $\cos \left( \theta +\beta \right)$ using the values of $\sin \left( \theta +\alpha \right)=a$ and $\sin \left( \theta +\beta \right)=b$ from the formulae $\cos x=\sqrt{1-{{\sin }^{2}}x}$. And after that we will find the value of $\cos \left( \alpha -\beta \right)=\cos \left[ \left( \theta +\alpha \right)-\left( \theta +\beta \right) \right]$ using $\cos \left( A-B \right)=\cos A\cos B+\sin A\sin B$ and substitute it in the equation we have $2{{\cos }^{2}}\left( \alpha -\beta \right)-1-4ab\cos \left( \alpha -\beta \right)$ and simplify it.

Complete step by step answer:
From the question we have that the value of $\sin \left( \theta +\alpha \right)=a$ and $\sin \left( \theta +\beta \right)=b$.
Since we know that $\cos x=\sqrt{1-{{\sin }^{2}}x}$ we can say that the value of $\cos \left( \theta +\alpha \right)=\sqrt{1-{{a}^{2}}}$ and$\cos \left( \theta +\beta \right)=\sqrt{1-{{b}^{2}}}$ .
Here we need the value of $\cos \left( \alpha -\beta \right)$ in order to answer the question. We can write $\alpha -\beta $ as $\left( \theta +\alpha \right)-\left( \theta +\beta \right)$ by adding and subtracting $\theta $ . Since we know that $\cos \left( A-B \right)=\cos A\cos B+\sin A\sin B$ . We can have the value of $\cos \left( \alpha -\beta \right)$ as $\cos \left( \alpha -\beta \right)=\cos \left[ \left( \theta +\alpha \right)-\left( \theta +\beta \right) \right]=\cos \left( \theta +\alpha \right)\cos \left( \theta +\beta \right)+\sin \left( \theta +\alpha \right)\sin \left( \theta +\beta \right)$ .
By substituting the respective values we will have $\sqrt{1-{{a}^{2}}}\sqrt{1-{{b}^{2}}}+ab$ .
By substituting this value $\cos \left( \alpha -\beta \right)=\sqrt{1-{{a}^{2}}}\sqrt{1-{{b}^{2}}}+ab$ in the place of the required value we will get $2{{\cos }^{2}}\left( \alpha -\beta \right)-1-4ab\cos \left( \alpha -\beta \right)=2{{\left( \sqrt{1-{{a}^{2}}}\sqrt{1-{{b}^{2}}}+ab \right)}^{2}}-1-4ab\left( \sqrt{1-{{a}^{2}}}\sqrt{1-{{b}^{2}}}+ab \right)$ .
By simplifying this equation we will have $2{{\left( \sqrt{1-{{a}^{2}}}\sqrt{1-{{b}^{2}}}+ab \right)}^{2}}-1-4{{a}^{2}}{{b}^{2}}-4ab\sqrt{1-{{a}^{2}}}\sqrt{1-{{b}^{2}}}$ .
Since we know that ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ by using this we will have $2\left( \left( 1-{{a}^{2}} \right)\left( 1-{{b}^{2}} \right)+2ab\left( \sqrt{1-{{a}^{2}}}\sqrt{1-{{b}^{2}}} \right)+{{a}^{2}}{{b}^{2}} \right)-1-4{{a}^{2}}{{b}^{2}}-4ab\sqrt{1-{{a}^{2}}}\sqrt{1-{{b}^{2}}}$ .
By simplifying this we will have $\begin{align}
  & 2\left( \left( 1-{{a}^{2}} \right)\left( 1-{{b}^{2}} \right) \right)+2{{a}^{2}}{{b}^{2}}+4ab\sqrt{1-{{a}^{2}}}\sqrt{1-{{b}^{2}}}-1-4{{a}^{2}}{{b}^{2}}-4ab\sqrt{1-{{a}^{2}}}\sqrt{1-{{b}^{2}}} \\
 & \Rightarrow 2\left( \left( 1-{{a}^{2}} \right)\left( 1-{{b}^{2}} \right) \right)-1-2{{a}^{2}}{{b}^{2}} \\
\end{align}$ .
By expanding this we will have $2\left( \left( 1-{{a}^{2}} \right)\left( 1-{{b}^{2}} \right) \right)-1-2{{a}^{2}}{{b}^{2}}\Rightarrow 2\left( 1-{{a}^{2}}-{{b}^{2}}+{{a}^{2}}{{b}^{2}} \right)-1-2{{a}^{2}}{{b}^{2}}$.
By expanding and simplifying this we will have $2-2{{a}^{2}}-2{{b}^{2}}+2{{a}^{2}}{{b}^{2}}-1-2{{a}^{2}}{{b}^{2}}=1-2{{a}^{2}}-2{{b}^{2}}$.
Hence we end up with the value that $2{{\cos }^{2}}\left( \alpha -\beta \right)-1-4ab\cos \left( \alpha -\beta \right)=1-2{{a}^{2}}-2{{b}^{2}}$ .
So now we have a conclusion that when $\sin \left( \theta +\alpha \right)=a$ and $\sin \left( \theta +\beta \right)=b$ , then $2{{\cos }^{2}}\left( \alpha -\beta \right)-1-4ab\cos \left( \alpha -\beta \right)=1-2{{a}^{2}}-2{{b}^{2}}$.

So, the correct answer is “Option B”.

Note: While answering of questions of this type we should remember that the formula is $\cos \left( A-B \right)=\cos A\cos B+\sin A\sin B$ not $\cos \left( A-B \right)=\cos A\cos B-\sin A\sin B$ . If we use this we will end up having a wrong answer.