Question

Tan theta+sec theta-1/tan theta-

sec theta+1=1+SinTheta/cos theta

Answer 1

Answer:

Hey Mate,

We have to prove LHS = RHS

LHS = $( tan\theta+sec\theta-1 ) / ( tan\theta-sec\theta+1 )$

RHS = $(sin\theta+1) / cos\theta$

Lets Start from LHS

LHS = $( tan\theta+sec\theta-1 ) / (tan\theta-sec\theta+1 )$

=  $( tan\theta+sec\theta-(sec^{2}\theta-tan^2\theta )) / ( tan\theta-sec\theta+1 )$

= $(tan\theta+sec\theta-[(sec\theta+tan\theta)(sec\theta-tan\theta)])/(tan\theta-sec\theta+1 )$

= $(tan\theta+sec\theta[tan\theta-sec\theta+1]) / (tan\theta-sec\theta+1 )$

= $tan\theta+sec\theta$

= $[sin\theta/cos\theta] + [1/cos\theta]$

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

Hence Proved,

LHS=RHS

Answer 2

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$\underline\bold{\huge{SOLUTION \: :}}$

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Here, i am holding the "THETAS" as "A".

So, we can proceed our solution as under :

$\boxed{L. H. S.}$

(tan A + sec A - 1)/(tan A - sec A + 1)

= (tan A + sec A - sec²A + tan²A)/(tan A - sec A + 1)

= [tan A + sec A - {(sec A+tan A) (sec A - tan A)}]/[tan A - sec A + 1]

= [tan A + sec A (1 - sec A + tan A)]/(tan A - sec A + 1)

= tan A + sec A

= sin A/cos A + 1/cos A

= ( 1 + sin A ) / cos A

= $\boxed{R. H. S.}$

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ULTIMATELY,

L.H.S. = R.H.S. (PROVED)

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