Prove that cot θ-cosθ/cotθ+cosθ=cosecθ-1/cosecθ+1

1.	Prove that cot θ-cosθ/cotθ+cosθ=cosecθ-1/cosecθ+1
Answer» ANSWER: Step-by-step EXPLANATION: $\Large{\underline{\rm{Given:}}}$ $\sf \dfrac{cot\theta-<klux>COS</klux>\theta}{cot\theta+cos\theta} =\dfrac{cosec\theta-1}{cosec\theta+1}$ $\Large{\underline{\rm{To\:Prove:}}}$ LHS = RHS $\Large{\underline{\rm{Identities\:used:}}}$ $\sf cot\theta=\dfrac{cos\theta}{sin\theta}$ $\sf \dfrac{1}{sin\theta} =cosec\theta$ $\Large{\underline{\rm{Proof:}}}$ ➞Here we have to the prove that the LHS of the equation = RHS. ➞ Taking the LHS of the equation, $\sf \dfrac{cot\theta-cos\theta}{cot\theta+cos\theta}$ ➞ Using identities, $\sf \implies \dfrac{\dfrac{cos\theta}{sin\theta}-cos\theta }{\dfrac{cos\theta}{sin\theta} +cos\theta}$ ➞ Taking cos θ COMMON from both numerator and denominator, $\sf \implies \dfrac{cos\theta(\dfrac{1}{sin\theta}-1) }{cos\theta(\dfrac{1}{sin\theta} +1)}$ ➞ Cancelling cos θ on both numerator and denominator, $\sf \implies \dfrac{(\dfrac{1}{sin\theta}-1) }{(\dfrac{1}{sin\theta} +1)}$ ➞ Using SUITABLE identities, $\sf \implies \dfrac{cosec\: \theta -1}{cosec\: \theta+1}$ $\implies \sf RHS$ ➞ Hence proved. $\Large{\underline{\rm{More\:Identities:}}}$ $\sf tan\theta=\dfrac{sin\theta}{cos\theta}$ $\sf \dfrac{1}{cos\theta} =sec\theta$ $\sf \dfrac{1}{tan\theta} =cot\theta$

Answer»

Step-by-step EXPLANATION:

$\Large{\underline{\rm{Given:}}}$