Question

<A+<B=90°, let us show that 1+tan A/ tam B = sec^2 A
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Answer 1

Answer:-

Given:-

∠A+∠B=90°

To find:-

1+tanA/tanB=sec²A

Solution:-

A+B=90°

B=90°-A

L.H.S=1+tanA/tanB

⇒1+tanA/tan(90°-A)    [tan( 90°-∅)=cot∅]

⇒1+tanA/cotA

⇒1+tan²A  [sec²∅-tan²∅=1⇒sec²∅=1+tan²∅]

⇒sec²A=R.H.S

Hence Proved.....