Question

prove :

(sec Φ - cos Φ)(cot Φ + tan Φ)
= tan Φ . sec Φ​

Answer 1

Answer:

$\displaystyle{(\sec\theta-\cos\theta)(\cot\theta+\tan\theta)=\tan.\sec \ [Proved]}$

Step-by-step explanation:

Given :

$\displaystyle{(\sec\theta-\cos\theta)(\cot\theta+\tan\theta)=\tan.\sec}\\\\\displaystyle{L.H.S.=(\sec\theta-\cos\theta)(\cot\theta+\tan\theta)}\\\\\displaystyle{\implies\left[\dfrac{1}{\cos\theta}-\cos\theta\right]\left[\dfrac{\cos\theta}{\sin\theta}+\frac{\sin\theta}{\cos\theta}\right]}\\\\\\\displaystyle{\implies\left[\dfrac{1-\cos^2\theta}{\cos\theta}\right]\left[\dfrac{\sin^2\theta+\cos^2\theta}{\sin\theta\cos\theta}\right]}$

$\displaystyle{\implies\left[\dfrac{\sin^2\theta}{\cos\theta}\right]\left[\dfrac{1}{\sin\theta\cos\theta}\right]}\\\\\\\displaystyle{\implies\left[\dfrac{\sin\theta\times\sin\theta}{\cos\theta}\times\dfrac{1}{\sin\theta\cos\theta}\right]}\\\\\\\displaystyle{\implies\left[\dfrac{\sin\theta}{\cos\theta}\right]\left[ \dfrac{1}{\cos\theta}\right]}\\\\\\\displaystyle{\implies\tan\theta \ .\sec\theta}$

L.H.S. =  R.H.S

Hence proved .

Answer 2

SOLUTION:-

Given:$(sec \theta - cos \theta)(cot \theta + tan \theta) = tan \theta \times sec \theta$

Take L.H.S

$(sec \theta - cos \theta)(cot \theta + tan \theta) \\ \\ = > ( \frac{1}{cos \theta} - cos \theta)( \frac{cos \theta}{sin \theta} + \frac{sin \theta}{cos \theta} ) \\ \\ = > ( \frac{1 - cos {}^{2} \theta }{cos \theta} )( \frac{cos {}^{2} \theta + {sin}^{2} \theta }{cos \theta \times sin \theta} ) \\ \\ = > ( \frac{ {sin}^{2} \theta }{cos \theta} )( \frac{1}{cos\theta \times sin \theta}) \\ \\ = > \frac{sin \theta}{cos \theta} \times \frac{1}{cos \theta} \\ \\ = > tan \theta \: \times sec \theta$

R.H.S

Hence, proved

Hope it helps ☺️