If A + B = 90^ , then + - ^2B ^2A is equal to : - Vidyadip

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 $(b)$

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "M.C.Q (1 Marks)" from Introduction of Trigonometry→Introduction of Trigonometry — all question types→Maths STD 10 question papers→CBSE Board STD 10 question papers→CBSE Board question paper generator→

Similar questions

If $P$ is a point on $x-$ axis such that its distance from the origin is $3$ units, then the coordinates of a point $Q$ on $OY $ such that $OP = OQ,$ are

For the following distribution:

Below:	10	20	30	40	50	60
Number of students:	3	12	27	57	75	80

the modal class is

Choose the correct answer from the given four options in the following questions : The quadratic equation $2\text{x}^2-\sqrt{5}\text{x}+1=0$ has.

The smallest number which when divided by 17, 23 and 29 leaves a remainder 11 in each case is

For the following distribution :

Mark	$60-70$	$70-80$	$80--90$	$90-100$	$100-110$
Frequency	$13$	$10$	$15$	$8$	$11$

The upper limit of the median class is :

The arithmetic mean of x, x + 3, x + 6, x + 9 and x + 12 is

The mean of the first $10$ prime numbers is :

Two dice are rolled together. The probability of getting sum of numbers on the two dice is 2, 3 or 5, is

If $5 \tan \theta-4=0$, then the value of $\frac{5 \sin \theta-4 \cos \theta}{5 \sin \theta+4 \cos \theta}$ is