Question

prove that tan(90-theta)+cot(90-theta)/cosec theta =sec A​

Answer 1

ANSWER:

To Prove:

• [tan(90-Θ) + cot(90-Θ)]/(cosecΘ) = secΘ

Proof:

$:\longrightarrow\dfrac{\tan(90-\theta)+\cot(90-\theta)}{\csc\theta}=\sec\theta\\\\\text{Solving LHS,}\\\\:\implies\dfrac{\tan(90-\theta)+\cot(90-\theta)}{\csc\theta}\\\\\text{We know that, \tan(90-A)=\cot A\:\:\&\:\:\cot(90-A)=\tan A . So,}\\\\:\implies\dfrac{\cot\theta+\tan\theta}{\csc\theta}\\\\\text{Also, \tan A=\dfrac{\sin A}{\cos A}, \cot A=\dfrac{\cos A}{\sin A}, \csc A=\dfrac{1}{\sin A} . So,}\\\\:\implies\dfrac{\dfrac{\sin\theta}{\cos\theta}+\dfrac{\cos\theta}{\sin\theta}}{\dfrac{1}{\sin\theta}}$

$:\implies \dfrac{\dfrac{\cos^2\theta+\sin^2\theta}{\cos\theta\sin\theta\!\!\!\!\!\!\!\!\!/}}{\dfrac{\:\:\:\:1\:\:\:}{\!\!\sin\theta\!\!\!\!\!\!\!\!\!/}}\\\\:\implies\dfrac{\cos^2\theta+\sin^2\theta}{\cos\theta}\\\\\text{We know that, \cos^2\theta+\sin^2\theta=1 . So,}\\\\:\implies\dfrac{1}{\cos\theta}\\\\\bf{:\implies\sec\theta}} = RHS\\\\\text{\bf{Hence Verified!!}}$

Formula Used:

• tan(90-Θ) = cotΘ
• cot(90-Θ) = tanΘ
• tanΘ = sinΘ/cosΘ
• cotΘ = cosΘ/sinΘ
• cosecΘ = 1/sinΘ
• sin²Θ+cos²Θ = 1