Question

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Answer 1

Answer:

Check the attachment

Hope you have got it

Answer 2

$\huge{\mathfrak{\underline{Answer:-}}}$

$\large{\underline{\mathfrak{Step \: By \: Step \: Explanation}}}$

Given:-

AB = a

BC = a

To find :-

(i) SinA

(ii) SecA

(iii) Cos²A + Sin²A

Solution :-

Apply Pythagoras theorem to Δ ABC

AC² = AB² + BC²

_________________[Put Values]

AC² = a² + a²

AC² = 2a²

AC = a√2

_____________________________________________

A.T.Q,

____________________

(i) SinA :-

SinA = Perpendicular/Hypotenuse

SinA = a/a√2

SinA = 1/√2 .............➊

______________________

(ii) SecA :-

» SecA = Hypotenuse/Base

» SecA = a√2 / a

» SecA = √2 ..............➋

________________________

(iii) Cos²A + Sin²A

$\rule{200}{2}$

We know that :-

SecA = 1/CosA

Put value of equation 2

Cos A = 1/√2 .........➌

$\rule{200}{2}$

Put values of equation 1 and 2

» (1/√2)² + (1/√2)²

» 1/2 + 1/2

» 2/2

» 1

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