Remember: We have to work from either the LHS or the RHS.
(Left hand side or the Right hand side)
You should already know this:
1.
\(\boxed{\bf{\huge{tan \theta = \frac{sin \theta}{cos \theta}}}}\)
2.
\(\boxed{\bf{\huge{cot\theta =\frac{1}{tan\theta}}}}\)
3.
\(\boxed{\bf{\huge{sin^2\theta +cos^2 \theta = 1}}}}\)
So, our question is:
\(\sf{\huge{tan\theta + cot\theta=\frac{1}{sin\theta cos\theta}}}\)
Plug in the first two identities I gave you.
\(\sf{\frac{sin\theta}{cos\theta} +\frac{1}{tan\theta} =\frac{1}{sin\theta cos\theta}\)
Apply the first identity I said you needed to know on 1/(tan θ). We should get:
\(\sf{\frac{sin\theta}{cos\theta} +\frac{1}{\frac{sin\theta}{cos\theta}} =\frac{1}{sin\theta cos\theta}}\\\\\\\sf{\frac{sin\theta}{cos\theta} +\frac{cos\theta}{sin\theta} =\frac{1}{sin\theta cos\theta}\)
Multiply the first fraction by sinθ, on both the numerator and denominator.
Multiply the second fraction by cosθ, on both the numerator and denominator.
\(\sf{\frac{sin\theta \times sin\theta}{cos\theta \times sin\theta} +\frac{cos\theta \times cos\theta}{sin\theta \times cos\theta} =\frac{1}{sin\theta cos\theta}}\\\\\\ \sf{\frac{sin^2\theta}{sin\theta cos\theta} + \frac{cos^2 \theta}{sin\theta cos\theta} = \frac{1}{sin\theta cos\theta}}\\\\\\\sf\frac{sin^2\theta + cos^2\theta}{sin\theta cos\theta} =\frac{1}{sin\theta cos\theta}}\)
Now, use the third identity I said that you needed to know to simplify the numerator.
\(\sf{\frac{sin^2\theta +cos^2\theta}{sin\theta cos\theta} =\frac{1}{sin\theta cos\theta}}\\\\\\\sf{\frac{1}{sin\theta cos\theta}=\frac{1}{sin\theta cos\theta}}\)
LHS = RHS
Therefore, identity is verified.
