Prove the following identites.
(i) If \(sin \ \theta + cos \ \theta = \sqrt{3}\), then prove that \(tan \ \theta + cot \ \theta = 1\)
Proof:
Consider \(sin \ \theta + cos \ \theta = \sqrt{3}\)
Squaring on both sides, we get:
\((sin \ \theta + cos \ \theta)^2 = (\sqrt{3})^2\)
By applying the formula \((a+b)^2=\)\(^2\)\(2ab\)\(^2\),
And the known trigonometric identity \(sin^2 \theta+cos^2 \theta=\) we get, \(sin \ \theta \ cos \ \theta = 1\)
(ii) If \(\sqrt{3} sin \ \theta - cos \ \theta = 0\), then show that \(tan \ 3 \theta = \frac{3 \ tan \ \theta - tan^3 \ \theta}{1 - 3 \ \tan^2 \ \theta}\)
Proof:
Consider \(\sqrt{3} sin \ \theta - cos \ \theta = 0\)
\(\sqrt{3} sin \ \theta = cos \ \theta\)
\(\frac{sin \ \theta}{cos \ \theta} = \frac{1}{\sqrt{3}}\)
\(tan \ \theta = \frac{1}{\sqrt{3}}\)
\(\theta =\)\(^{\circ}\)
Now, consider LHS.
LHS \(= tan \ 3 \theta\)
\(= tan \ 3(\)\(^{\circ})\)
\(= tan\)\(^{\circ}\)
\(= \text{undefined}\) ---- (\(1\))
Now consider RHS.
RHS \(= \frac{3 \ tan \ \theta - tan^3 \ \theta}{1 - 3 \ \tan^2 \ \theta}\)
\(= \frac{3 \ tan \ 30^{\circ} - tan^3 \ 30^{\circ}}{1 - 3 \ tan^2 \ 30^{\circ}}\)
After simplyfing this we get,
\(RHS= \text{undefined}\) ---- (\(2\))
From equations (\(1\)) and (\(2\)), we can see that LHS \(=\) RHS.