Is csc(x)+cot(x)sec(x)-1=tan(x) ?

The answer to whether csc(x)+cot(x)sec(x)-1=tan(x) is False

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prove csc(x)+cot(x)sec(x)-1=tan(x)

prove

csc (x) + cot (x) sec (x) - 1 = tan (x)

False

Solution steps

csc (x) + cot (x) sec (x) - 1 = tan (x)

Show that the two sides are not equal

For

csc (x) + cot (x) sec (x) - 1 = tan (x)

plug in

x = 1

csc (1) + cot (1) sec (1) - 1 = 1.37679 \dots

csc (1) + cot (1) sec (1) - 1

Refine to a decimal form

= 1.37679 \dots

tan (1) = 1.55740 \dots

tan (1)

Refine to a decimal form

= 1.55740 \dots

The two sides are not equal

\Rightarrow False

\frac{3 csc ( x ) - 3 sin ( x )}{tan ( x ) - cot ( x )} = 3 cos^{3} (x)

9 cos (x^{\circ}) + 6 sin (x^{\circ}) = 10

\frac{tan ^{2} ( A )}{sec ^{2} ( A )} = sin^{2} (A)

sin (2 x) - cot (x) = cot (x) cos (2 x)

tan (x) + cot (x) \cdot tan (x) = sec^{2} (x)

Is csc(x)+cot(x)sec(x)-1=tan(x) ?
The answer to whether csc(x)+cot(x)sec(x)-1=tan(x) is False