Question

Given cos A= 5/13 evaluate sec²A-tan A/2cotA​

Answer 1

If cos A = 5/13 means Base = 5 , and HYPOTENUSE = 13.

by Pythagoras Theorem :-

P = √H^2 - B^2

P = √169 - 25

P = √144 = 12

so , Sec^2 A = (H/B)^2 = (13/5)^2 = 169/25

Tan A = P/B = 12/5

Cot A = B/P = 5/12

so , solving the above Equation :-

169/25 - 12/5 /2×5/12

= 169-60/25/5/6

= 100/25/5/6

= 4/5 ×6 = 24/5

Answer 2

Cos A = 5/13

To find :

Sec^2 A - tan A/ 2 cot A

Solve :

Cos A = Base/Hypotenuse

B/H = 5/13

B = 5

H = 13

H^2 = B^2 + P^2

13^2 = 5^2 + P^2

P^2 = 169 +25 = 144

P = 12

As we know that,

Sec Φ = H/B

tan Φ = P/B

Cot Φ = B/P

Here, we have,

H = 13

B = 5

P = 12

So,

Sec^2 A - tan A/ 2cot A

=> (13^2/5^2) - (12/5)/2(5/12)

=> (169/25) - (12/5) / (10/12)

=> (169/25) - 12/5 x 12/10

=> (169/25) - 144/50

=> 24/5

Ans. = 24/5