MCQ
$[(\sec A+\tan A)(1-\sin A)]$ on simplification gives
- A$\tan ^2 A$
- B$\sec ^2 A$
- ✓$\cos A$
- D$\sin A$
✓
Answer
Correct option: C.
$\cos A$
Using $\sec A=\frac{1}{\cos A}$ and $\tan A=\frac{\sin A}{\cos A}$ in given expression, we get
$[(\sec A+\tan A)(1-\sin A)]=\left[\begin{array}{l}
\left( \frac{1}{\cos A}+\frac{\sin A}{\cos A}\right) \\
(1-\sin A)
\end{array}\right].$
Apply $(a+b)(a-b)=a^2-b^2$ where $a=1$ and $b=\sin A$
$\Rightarrow[(\sec A+\tan A)(1-\sin A)]=\frac{1-\sin ^2 A}{\cos A}\quad \quad \ldots \ldots(i) .$
Also, $1-\sin ^2 A=\cos ^2 A\quad \quad \ldots \ldots(ii)$
Substitute equation $(ii)$ in equation $(i),$
$\Rightarrow[(\sec A+\tan A)(1-\sin A)]=\frac{\cos ^2 A}{\cos A}=\cos A$
Hence, the correct option is $(c).$
$[(\sec A+\tan A)(1-\sin A)]=\left[\begin{array}{l}
\left( \frac{1}{\cos A}+\frac{\sin A}{\cos A}\right) \\
(1-\sin A)
\end{array}\right].$
Apply $(a+b)(a-b)=a^2-b^2$ where $a=1$ and $b=\sin A$
$\Rightarrow[(\sec A+\tan A)(1-\sin A)]=\frac{1-\sin ^2 A}{\cos A}\quad \quad \ldots \ldots(i) .$
Also, $1-\sin ^2 A=\cos ^2 A\quad \quad \ldots \ldots(ii)$
Substitute equation $(ii)$ in equation $(i),$
$\Rightarrow[(\sec A+\tan A)(1-\sin A)]=\frac{\cos ^2 A}{\cos A}=\cos A$
Hence, the correct option is $(c).$
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