Question

prove that tan A + sec Aminus 1 upon tan A - sec A+ 1 equal to 1 + sin a upon Cos A​

Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Answer:

(tanA+ secA-1)/(tanA-secA+1)=secA +tanA [so we have to prove this equation]

First, add 1 on both sides,

or,(tanA + secA -1)/(tanA-secA+1)+1=secA + tanA+1

Then on simplifying you'll get this…[Trust me it's a one line simplication]

or, 2tanA/(tanA-secA+1)=tanA+secA+1

Then group them as (tanA+1) and secA and cross multiply…

2tanA={(tanA+1)+secA}{(tanA+1)-secA}

or, 2tanA=(tanA+1)² - sec²A…….(i)

Yes , I have used that a²-b² formula…..

Now you can prove the LHS and RHS..

RHS:

Simplify that part (i)…

=tan²A +2tanA-sec²A+1

Now, tan²A-sec²A=-1…so, substituting the value…

=2tanA+1–1

=2tanA

Hence LHS AND RHS proved which ultimately solves your question…