Question

Choose the correct option. Justify your choice. $\left(1+\tan \theta +\sec \theta \right)\left(1+\cot \theta \mathit{-}\mathrm{cosec}\theta \right)=$

• A. $0$
• B. $1$
• C. $2$
• D. $-1$

Answer 1

The correct option is C

$2$

Determine the correct option.

The explanation for the correct option:

We know that,

$$\begin{gathered} \tan \theta =\frac{\sin \theta }{\cos \theta } \\ \sec \theta =\frac{1}{\cos \theta } \\ \cot \theta =\frac{\cos \theta }{\sin \theta } \\ \mathrm{cosec}\theta =\frac{1}{\sin \theta } \end{gathered}$$

Substitute the above values in the given equation and simplify:

$\begin{array}{rcl}\left(1+\tan \theta +\sec \theta \right)\left(1+\cot \theta \mathit{-}\mathrm{cosec}\theta \right)& =& \left(1+\frac{\sin \theta }{\cos \theta }+\frac{1}{\cos \theta }\right)\left(1+\frac{\cos \theta }{\sin \theta }-\frac{1}{\sin \theta }\right)\\ & =& \left(\frac{\cos \theta +\sin \theta +1}{\cos \theta }\right)\left(\frac{\sin \theta +\cos \theta -1}{\sin \theta }\right)\\ & =& \frac{\left\{\left(\cos \theta +\sin \theta \right)+1\right\}\left\{\left(\sin \theta +\cos \theta \right)-1\right\}}{\cos \theta \sin \theta }\\ & =& \frac{{\left(\cos \theta +\sin \theta \right)}^2-1^2}{\cos \theta \sin \theta }\left[\because \left(a+b\right)\left(a-b\right)=a^2-b^2\right]\\ & =& \frac{{\cos }^2\theta +{\sin }^2\theta +2\cos \theta \sin \theta -1}{\cos \theta \sin \theta }\\ & =& \frac{1+2\cos \theta \sin \theta -1}{\cos \theta \sin \theta }\left[\because {\cos }^2\theta +{\sin }^2\theta =1\right]\\ & =& \frac{2\cos \theta \sin \theta }{\cos \theta \sin \theta }\\ & =& 2\end{array}$

Therefore, $\left(1+\tan \theta +\sec \theta \right)\left(1+\cot \theta \mathit{-}\mathrm{cosec}\theta \right)=2$.

The explanation for the incorrect option:

Since the value determined above$\left(1+\tan \theta +\sec \theta \right)\left(1+\cot \theta \mathit{-}\mathrm{cosec}\theta \right)=2$ is not equivalent to the value mentioned in Option A, Option B, and Option D.

Hence,Option A, Option B, and Option D are incorrect option

Hence, the correct option is (C) i.e., $2$.