Question

trigonometric problem​

Answer 1

Question:

Prove that:

$\sf\:1\:+\:\dfrac{\cot^2\:\alpha}{1\:+\:\text{cosec}\sf\:\alpha}\:=\:\text{cosec}\:\alpha$

Answer:

$\boxed{\red{\sf\:1\:+\:\dfrac{\cot^2\:\alpha}{1\:+\:\text{cosec}\sf\:\alpha}\:=\:\text{cosec}\sf\:\alpha}}$

Step-by-step-explanation:

We have to prove LHS = RHS in the given trigonometric equation.

We can prove the given trigonometric equation by using LHS.

$\displaystyle\sf\:LHS\:=\:1\:+\:\dfrac{\cot^2\:\alpha}{1\:+\:\text{cosec}\sf\:\alpha}\\\\\\\sf\:LHS\:=\:\dfrac{1\:+\:\text{cosec}\:\alpha\:+\:\cot^2\:\alpha}{1\:+\:\text{cosec}\:\alpha}\\\\\\\sf\:LHS\:=\:\dfrac{\text{cosec}\:\alpha\:+\:\underline{1\:+\:\cot^2\:\alpha}}{1\:+\:\text{cosec}\:\alpha}\\\\\\\sf\:LHS\:=\:\dfrac{\text{cosec} \:\alpha\:+\:\text{cosec}^{\sf\:2}\:\alpha}{1\:+\:\text{cosec}\:\alpha}\:\:\:-\:-\:[\:\because\:1\:+\:\cot^2\:\alpha\:=\:\text{cosec}^{\sf\:2}\:\alpha\:]\\\\\\\sf\:LHS\:=\:\dfrac{\text{cosec}\:\alpha\:\cancel{(\:1\:+\:\text{cosec}\:\alpha\:)}}{\cancel{(\:1\:+\:\text{cosec}\:\alpha\:)}}\\\\\\\sf\:LHS\:=\:\text{cosec}\:\alpha\\\\\\\boxed{\red{\sf\:LHS\:=\:RHS}}$

Hence proved!

$\rule{200}{1}$

$\boxed{\red{\bf\:Alternative\:method}}$

We can prove the given trigonometric equation by using RHS.

$\displaystyle\sf\:RHS\:=\:\text{cosec}\:\alpha\\\\\\\sf\:RHS\:=\:\text{cosec}\:\alpha\:\times\:\dfrac{\:(\:1\:+\:\text{cosec}\:\alpha\:)}{\:(\:1\:+\:\text{cosec}\:\alpha\:)\:}\:\:\:-\:-\:[\:Multiplying\:\&\:dividing\:by\:(\:1\:+\:\text{cosec}\:\alpha\:)\:]\\\\\\\sf\:RHS\:=\:\dfrac{\text{cosec}\:\alpha\:+\:\underline{\text{cosec}^{\sf\:2}\:\alpha}}{1\:+\:\text{cosec}\:\alpha}\\\\\\\sf\:RHS\:=\:\dfrac{\text{cosec}\:\alpha\:+\:1\:+\:\cot^2\:\alpha}{1\:+\:\text{cosec}\:\alpha}\:\:\:-\:-\:[\:\because\:\text{cosec}^{\sf\:2}\:\alpha\:=\:1\:+\:\cot^2\:\alpha\:]\\\\\\\sf\:RHS\:=\:\cancel{\dfrac{1\:+\:\text{cosec}\:\alpha}{1\:+\:\text{cosec}\:\alpha}}\:+\:\dfrac{\cot^2\:\alpha}{1\:+\:\text{cosec}\:\alpha}\:\:\:-\:-\:[\:Separating\:the\:denominator\:]\\\\\\\sf\:RHS\:=\:1\:+\:\dfrac{\cot^2\:\alpha}{1\:+\:\text{cosec}\:\alpha}\\\\\\\boxed{\red{\sf\:RHS\:=\:LHS}}$

Hence proved!

$\rule{200}{1}$

Additional Information:

$\displaystyle\boxed{\begin{minipage}{5 cm}\underline{\bf\:Trigonometric\:Identities}\\\\\sf\:1.\:\tan\:\theta\:=\:\dfrac{\sin\:\theta}{\cos\:\theta}\\\\\sf\:2.\:\sin^2\:\theta\:+\:\cos^2\:\theta\:=\:1\\\\\sf\:3.\:1\:+\:\cot^2\:\theta\:=\:\text{cosec}^{\sf\:2}\:\theta\\\\\sf\:4.\:1\:+\:\tan^2\:\theta\:=\:\sec^2\:\theta\end{minipage}}$