Question

tan theta+sec theta-1/tan theta-sec theta+1=1+sin theta/cos theta ​

Answer 1

Step-by-step explanation:

Hey Mate,

We have to prove LHS = RHS

LHS = ( tan\theta+sec\theta-1 ) / ( tan\theta-sec\theta+1 )(tanθ+secθ−1)/(tanθ−secθ+1)

RHS = (sin\theta+1) / cos\theta(sinθ+1)/cosθ

Lets Start from LHS

LHS = ( tan\theta+sec\theta-1 ) / (tan\theta-sec\theta+1 )(tanθ+secθ−1)/(tanθ−secθ+1)

= ( tan\theta+sec\theta-(sec^{2}\theta-tan^2\theta )) / ( tan\theta-sec\theta+1 )(tanθ+secθ−(sec

2

θ−tan

2

θ))/(tanθ−secθ+1)

= (tan\theta+sec\theta-[(sec\theta+tan\theta)(sec\theta-tan\theta)])/(tan\theta-sec\theta+1 )(tanθ+secθ−[(secθ+tanθ)(secθ−tanθ)])/(tanθ−secθ+1)

= (tan\theta+sec\theta[tan\theta-sec\theta+1]) / (tan\theta-sec\theta+1 )(tanθ+secθ[tanθ−secθ+1])/(tanθ−secθ+1)

= tan\theta+sec\thetatanθ+secθ

= [sin\theta/cos\theta] + [1/cos\theta][sinθ/cosθ]+[1/cosθ]

= [(sin\theta+1) /cos\theta][(sinθ+1)/cosθ] = RHS

Hence Proved,

LHS=RHS

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Answer 2

Answer:

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