Question
Prove that:
(sec θ – cos θ) (cot θ + tan θ) = tan θ sec θ
(sec θ – cos θ) (cot θ + tan θ) = tan θ sec θ
✓
Answer
Taking LHS
$\begin{array}{l}(\sec \theta-\cos \theta)(\cot \theta+\tan \theta) \\
=\sec \theta \cot \theta+\sec \theta \tan \theta-\cos \theta \cot \theta-\cos \theta \tan \theta \\
=\frac{1}{\cos \theta}\left(\frac{\cos \theta}{\sin \theta}\right)+\sec \theta \tan \theta-\frac{\cos \theta(\cos \theta)}{\sin \theta}-\cos \theta\left(\frac{\sin \theta}{\cos \theta}\right) \\
=\frac{1}{\sin \theta}-\frac{\cos ^2 \theta}{\sin \theta}-\sin \theta+\tan \theta \sec \theta
\end{array}$
Taking LCM of first three terms,
$\begin{array}{l}
=\frac{1-\cos ^2 \theta-\sin ^2 \theta}{\sin \theta}+\tan \theta \sec \theta \\
=\frac{1-\left(\cos ^2 \theta+\sin ^2 \theta\right)}{\sin \theta}+\tan \theta \sec \theta \\
=\frac{1-1}{\sin \theta}+\tan \theta \sec \theta\left[\text { As, } \sin ^2 \theta+\cos ^2 \theta=1\right] \\
=\tan \theta \sec \theta \\
=\text { RHS }\end{array}$
Proved!
$\begin{array}{l}(\sec \theta-\cos \theta)(\cot \theta+\tan \theta) \\
=\sec \theta \cot \theta+\sec \theta \tan \theta-\cos \theta \cot \theta-\cos \theta \tan \theta \\
=\frac{1}{\cos \theta}\left(\frac{\cos \theta}{\sin \theta}\right)+\sec \theta \tan \theta-\frac{\cos \theta(\cos \theta)}{\sin \theta}-\cos \theta\left(\frac{\sin \theta}{\cos \theta}\right) \\
=\frac{1}{\sin \theta}-\frac{\cos ^2 \theta}{\sin \theta}-\sin \theta+\tan \theta \sec \theta
\end{array}$
Taking LCM of first three terms,
$\begin{array}{l}
=\frac{1-\cos ^2 \theta-\sin ^2 \theta}{\sin \theta}+\tan \theta \sec \theta \\
=\frac{1-\left(\cos ^2 \theta+\sin ^2 \theta\right)}{\sin \theta}+\tan \theta \sec \theta \\
=\frac{1-1}{\sin \theta}+\tan \theta \sec \theta\left[\text { As, } \sin ^2 \theta+\cos ^2 \theta=1\right] \\
=\tan \theta \sec \theta \\
=\text { RHS }\end{array}$
Proved!
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