Question

Tan(x)+cot(x)=sec(x)csc(x)

Answer 1

The LHS is equal to the RHS, and the trigonometric identity is proved:

`\[ \tan(x) + \cot(x) = \sec(x) \csc(x) \]`

To prove the trigonometric identity
`\( \tan(x) + \cot(x) = \sec(x) \csc(x) \),`we'll start with the left-hand side (LHS) and manipulate it to match the right-hand side (RHS).

Starting with the LHS:

`\[ \tan(x) + \cot(x) \]`

We know that
`\( \tan(x) = (\sin(x))/(\cos(x)) \)` and
`\( \cot(x) = (1)/(\tan(x)) = (\cos(x))/(\sin(x)) \).`

Now, substitute these expressions into the LHS:

`\[ (\sin(x))/(\cos(x)) + (\cos(x))/(\sin(x)) \]`

To combine these fractions, find a common denominator, which is the product of the denominators
`(\( \cos(x) * \sin(x) \))`

`\[ (\sin^2(x) + \cos^2(x))/(\cos(x) \sin(x)) \]`

Recall that
`\( \sin^2(x) + \cos^2(x) = 1 \) (the Pythagorean identity),` so the expression becomes:

`\[ (1)/(\cos(x) \sin(x)) \]`

Now, express
`\( (1)/(\cos(x) \sin(x)) \) in terms of sec(x) and csc(x)`:

`\[ (1)/(\cos(x) \sin(x)) = (1)/(\cos(x)) \cdot (1)/(\sin(x)) = \sec(x) \csc(x) \]`

Therefore, we've shown that the LHS is equal to the RHS, and the trigonometric identity is proved:

`\[ \tan(x) + \cot(x) = \sec(x) \csc(x) \]`