What is the value of cos{cos-1(3/5) + cos-1(12/13)}1. 33/652. 20/653.

1.	What is the value of cos{cos-1(3/5) + cos-1(12/13)}1. 33/652. 20/653. 16/654. 36/65
Answer» Correct Answer - Option 3 : 16/65 Concept: cos(A + B) = cos A cos B - sin A sin B \(\rm sin^2x+cos^2x=1\) Calculation: To find: cos{cos-1(3/5) + cos-1(12/13)} = ? Let, cos-1(3/5) = A ⇒cos A = 3/5 ⇒ sin A = \(\sqrt{({1-\frac{3^2}{5^2}})}\) = √ (16/25) = 4/5 Also, let cos-1(12/13) = B ⇒cos B = 12/13 ⇒ sin B = \(\sqrt{(1-\frac{12^2}{13^2})}=\sqrt{(\frac{169-144}{169})}\) = √(25/169) = 5/13 A + B = cos-1(3/5) + cos-1(12/13) cos{cos-1(3/5) + cos-1(12/13)} = cos(A + B) Now, we know, cos(A + B) = cos A cos B - sin A sin B \(=\frac{3}{5}\times \frac{12}{13}-\frac{4}{5}\times \frac{5}{13}\\ =\frac{36}{65}-\frac{20}{65}\) = 16/65 Hence, option (3) is correct.

What is the value of cos{cos-1(3/5) + cos-1(12/13)}1. 33/652. 20/653. 16/654. 36/65

Answer» Correct Answer - Option 3 : 16/65

Concept:

cos(A + B) = cos A cos B - sin A sin B

\(\rm sin^2x+cos^2x=1\)

Calculation:

To find: cos{cos-1(3/5) + cos-1(12/13)} = ?

Let, cos-1(3/5) = A ⇒cos A = 3/5

⇒ sin A = \(\sqrt{({1-\frac{3^2}{5^2}})}\)

= √ (16/25)

= 4/5

Also, let cos-1(12/13) = B ⇒cos B = 12/13

⇒ sin B = \(\sqrt{(1-\frac{12^2}{13^2})}=\sqrt{(\frac{169-144}{169})}\)

= √(25/169)

= 5/13

A + B = cos-1(3/5) + cos-1(12/13)

cos{cos-1(3/5) + cos-1(12/13)} = cos(A + B)

Now, we know, cos(A + B) = cos A cos B - sin A sin B

\(=\frac{3}{5}\times \frac{12}{13}-\frac{4}{5}\times \frac{5}{13}\\ =\frac{36}{65}-\frac{20}{65}\)

= 16/65

Hence, option (3) is correct.