Trigonometric Functions — MATHS STD 11 Science — Question

2. -1

Solution:

$\pi=180^\circ$

$\sec\text{A}(\cos\text{B}\cos\text{C}-\sin^2\text{B}\sin\text{C})=\frac{\cos\text{B}\cos(\pi-(\text{A+B}))-\sin\text{B}\sin(\pi(\text{A+B}))}{\cos\text{A}}$

We know that, $\cos(\pi-\theta)=-\cos\theta$ and $\sin(\pi-\theta)=\sin\theta,$

$\therefore\sec\text{A}(\cos\text{B}\cos\text{C}-\sin\text{B}\sin\text{C})=\frac{\cos\text{B}\cos(\text{A+B})-\sin\text{B}\sin(\text{A+B})}{\cos\text{A}}$

Now, using the identities $\cos(\text{A+B})=\cos\text{A}\cos\text{B}-\sin\text{A}\sin\text{B}$ and $\sin(\text{A+B})=\sin\text{A}\cos\text{B}+\cos\text{A}\sin\text{B},$ we get

$\sec\text{A}(\cos\text{B}\cos\text{C}-\sin\text{B}\sin\text{C})=\frac{-\cos\text{A}\cos\text{B}^2+\cos\text{B}\sin\text{A}\sin\text{B}-\sin\text{B}\sin\text{A}\cos\text{B}-\sin^2\cos\text{A}}{\cos\text{A}}$

$\Rightarrow\sec\text{A}(\cos\text{B}\cos\text{C}-\sin\text{B}\sin\text{C})=\frac{-\cos\text{A}(\cos^2\text{B}+\sin^2\text{B})}{\cos\text{A}}$

$\Rightarrow\sec\text{A}(\cos\text{B}\cos\text{C}-\sin\text{B}\sin\text{C})=\frac{-\cos\text{A}}{\cos\text{A}}=-1$