Question

please answer fast very urgent!!
If K - 7 sec^2 62° + 7 cot^2 28° = 7 sec 0° then find the value of K

Answer 1

Answer:

k = 0

Step-by-step explanation:

Given :

k - 7sec²62° + 7 cot²28° = 7sec 0°

=> k - 7sec²62° + 7 Cot² (90 - 62)° = 7sec 0°

=> k - 7 (sec²62° - Tan²62°) = 7 Sec0°  

(∵Sec²θ - Tan²θ = 1 and Sec0° = 1)

=> k + 7(1) = 7 (1)

=> k + 7 = 7

=> k = 0.

Hope it helps!Brainliest it Bhai!

Answer 2

Step-by-step explanation:

K-7sec^2 62°+7cot^2 28°=7sec0°

K-[(7)(tan^2 62°+1)]-7cot^2 28°=7(1)

K-[(7){(cot^2 28°+1-7cot^2 28°}]=7

K-[7cot^2 28°+7-7cot^2 28°]=7

K-[7]=7

K=7+7

K=14