please answer fast very urgent!!
If K - 7 sec^2 62° + 7 cot^2 28° = 7 sec 0° then find the value of K
Answers
Answered by
2
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!
Answered by
1
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
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