Question

____________________??

Answer 1

$\huge{\mathcal{\underline{\green{SOLUTION}}}}$

Let,

$x = {tan}^{ - 1} \sqrt{3}$

$\implies \: tanx \: = \sqrt{3}$

Now the range of principal value of

${tan}^{ - 1} x \: \: is \: \: ( - \frac{\pi}{2} , \frac{\pi}{2} )$

Hence

${tan}^{ - 1} \sqrt{3} = \frac{\pi}{3}$

Again let

$y = {sec}^{ - 1} ( - 2)$

So

$\implies \: sec \: y = - 2$

Now the principal value of

${sec}^{ - 1} x \: \: is \: \: [0 ,\pi ] - \frac{\pi}{2}$

Hence

${sec}^{ - 1} ( - 2) = \frac{2\pi}{3}$

Therefore

${tan}^{ - 1} \sqrt{3} \: - \: {sec}^{ - 1} ( - 2) = \frac{\pi}{3} - \frac{2\pi}{3} = - \frac{\pi}{3}$

Answer 2

Answer:

Let x= cos-1(12/13)

Cosx =12/13

Sinx = root over 1-cos²x

=Root over 1-(12/13)²

=root over 1-144/169

=root over 25/169

Sinx=5/13

Again let y= sin -¹3/5

Sin y=3/5

Cos y=root over 1-sin²y

=root over 1-(3/5)²

=root over 1-9/25

=root over 16/25

Cos y = 4/5

Sin (x+y)= sinxcosy + cosxsiny

Sin(x+y)= 5/13×4/5+12/13×3/5

20/65 + 36/65

56/65

Sin(x+y) = 56/65

x+y= sin-¹ 56/65

cos-¹(12/13) + sin-¹(3/5) = sin-¹(56/65)

So the answer is option d