Question

prove that cot θ-cosθ/cotθ+cosθ=cosecθ-1/cosecθ+1​

Answer 1

Answer:

Formula:

cotθ=

sinθ

cosθ

=cosθcosecθ

LHS=

cotθ+cosθ

cotθ−cosθ

=

cosθcosecθ−cosθ

cosθcosecθ−cosθ

=

cosθ(cosecθ−1)

cosθ(cosecθ−1)

=

cosecθ+1

cosecθ−1

=RHS

Hence proved.

Hope it helps!!

Answer 2

Answer:

Step-by-step explanation:

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

$\sf \dfrac{cot\theta-cos\theta}{cot\theta+cos\theta} =\dfrac{cosec\theta-1}{cosec\theta+1}$

$\Large{\underline{\rm{To\:Prove:}}}$

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.

$\Large{\underline{\rm{More\:Identities:}}}$

$\sf tan\theta=\dfrac{sin\theta}{cos\theta}$

$\sf \dfrac{1}{cos\theta} =sec\theta$

$\sf \dfrac{1}{tan\theta} =cot\theta$