Question

prove that sec^2=1+tan^2​

Answer 1

Answer:

this is your answer

Step-by-step explanation:

hope it's help

Answer 2

Answer:

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Step-by-step explanation:

Let ΔABP is a right angle triangle ∠B=90

o

. Let ∠PAB=θ.

By Pythagorus theorem,

In ΔABP,

PA 2 =PB 2+AB 2

Dividing both sides of equation (i) by AB

2

⇒(secθ) 2 =(tanθ) 2 +1

∵tanθ= AB/PB

secθ= AB/ PB

⇒sec 2 θ=tan 2 θ+1.