Question

Sinx - cosx =0 find the value of secx

Answer 1

Answer:

$\displaystyle \sec x=\pm\sqrt{2}$

Step-by-step explanation:

Trigonometric Equations

We need to recall the identity:

$\sin^2 x+\cos ^2 x=1$

Solving for the sine:

$\sin^2 x=1-\cos ^2 \qquad\qquad [1]$

We are given the equation:

$\sin x-\cos x=0$

Or, equivalently:

$\sin x=\cos x$

Squaring:

$\sin^2 x=\cos^2 x$

Substituting from [1]

$1-\cos ^2 =\cos^2 x$

Simplifying:

$2\cos ^2=1$

Solving:

$\displaystyle \cos^2x=\frac{1}{2}$

$\displaystyle \cos x=\pm\sqrt{\frac{1}{2}}$

Since:

$\sec x = \frac{1}{\cos x}:$

$\mathbf{\displaystyle \sec x=\pm\sqrt{2}}$