MCQ
If $k+1=\sec ^2 \theta(1+\sin \theta)(1-\sin \theta)$, then $k=$
- A1
- B2
- ✓$0$
- D-1
✓
Answer
Correct option: C.
$0$
(C)0
We have,
$
\begin{array}{l}
k+1=\sec ^2 \theta(1+\sin \theta)(1-\sin \theta) \\
\Rightarrow \quad k+1=\frac{1}{\cos ^2 \theta} \times\left(1-\sin ^2 \theta\right) \Rightarrow k+1=\frac{\cos ^2 \theta}{\cos ^2 \theta} \Rightarrow k+1=1 \Rightarrow k=0
\end{array}
$
We have,
$
\begin{array}{l}
k+1=\sec ^2 \theta(1+\sin \theta)(1-\sin \theta) \\
\Rightarrow \quad k+1=\frac{1}{\cos ^2 \theta} \times\left(1-\sin ^2 \theta\right) \Rightarrow k+1=\frac{\cos ^2 \theta}{\cos ^2 \theta} \Rightarrow k+1=1 \Rightarrow k=0
\end{array}
$
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