Question

If k+7 sec^262-7 cot^228=7 sec 0 find the value of k

Answer 1

The value of k is zero(0).

Step-by-step explanation:

Given,

$k+7\sec^262-7\cot^228=7\sec 0$

To find, the value of k =

∴ $k+7\sec^262-7\cot^228=7\sec 0$

⇒ $k+7\sec^262-7\cot^2(90-62)=7\sec 0$

Using trigonometric identity,

$\tan A=\cot (90-A)$

⇒ $k+7(\sec^262-\tan^262)=7\sec 0$

Using trigonometric identity,

$\sec^2A-\tan^2A=1$

⇒$k+7(1)=7(1)$

⇒ k + 7 = 7

⇒  k = 7 - 7 = 0

Hence, the value of k is zero(0).