MCQ
If $A + B = 90^\circ$ , then $\frac{\tan\text{A}\tan\text{B}+\tan\text{A}\cot\text{B}}{\sin\text{A}\sec\text{B}}-\frac{\sin^2\text{B}}{\cos^2\text{A}}$ is equal to:
- A$\cot^2\text{A}$
- ✓$\cot^2\text{B}$
- C$-\tan^2\text{A}$
- D$-\cot^2\text{A}$
✓
Answer
Correct option: B.
$\cot^2\text{B}$
We have:
$\text{A+B}=90^\circ$
$\Rightarrow\text{B}=90^\circ-\text{A}$
We have to find the value of the following expression
$\frac{\tan\text{A}\tan\text{B}+\tan\text{A}\cot\text{B}}{\sin{\text{A}\sec\text{B}}}-\frac{\sin^2\text{B}}{\cos^2\text{A}}$
So,
$\frac{\tan\text{A}\tan\text{B}+\tan\text{A}\cot\text{B}}{\sin{\text{A}\sec\text{B}}}-\frac{\sin^2\text{B}}{\cos^2\text{A}}$
$=\frac{\tan\text{A}\tan(90^\circ-\text{A})+\tan\text{A}\cot(90^\circ-\text{A})}{\sin{\text{A}\sec(90^\circ-\text{A})}}-\frac{\sin^2(90^\circ-\text{A})}{\cos^2\text{A}}$
$=\frac{\tan\text{A}\cot{\text{A}}+\tan\text{A}\tan\text{A}}{\sin\text{A}\text{cosecA}}-\frac{\cos^2\text{A}}{\cos^2\text{A}}$
$=1+\tan^2\text{A}-1$
$=\tan^2\text{A}$
$=\tan^2(90^\circ-\text{B})$
$=\cot^2\text{B}$
Hence the correct option is $(b)$
$\text{A+B}=90^\circ$
$\Rightarrow\text{B}=90^\circ-\text{A}$
We have to find the value of the following expression
$\frac{\tan\text{A}\tan\text{B}+\tan\text{A}\cot\text{B}}{\sin{\text{A}\sec\text{B}}}-\frac{\sin^2\text{B}}{\cos^2\text{A}}$
So,
$\frac{\tan\text{A}\tan\text{B}+\tan\text{A}\cot\text{B}}{\sin{\text{A}\sec\text{B}}}-\frac{\sin^2\text{B}}{\cos^2\text{A}}$
$=\frac{\tan\text{A}\tan(90^\circ-\text{A})+\tan\text{A}\cot(90^\circ-\text{A})}{\sin{\text{A}\sec(90^\circ-\text{A})}}-\frac{\sin^2(90^\circ-\text{A})}{\cos^2\text{A}}$
$=\frac{\tan\text{A}\cot{\text{A}}+\tan\text{A}\tan\text{A}}{\sin\text{A}\text{cosecA}}-\frac{\cos^2\text{A}}{\cos^2\text{A}}$
$=1+\tan^2\text{A}-1$
$=\tan^2\text{A}$
$=\tan^2(90^\circ-\text{B})$
$=\cot^2\text{B}$
Hence the correct option is $(b)$
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