Question

1. From the following figure, find (sec xº + tan xº)​

Answer 1

Answer:

Mark me as brainliest

Step-by-step explanation:

Let the given triangle be ABC, where ∠ABC=90

∘

, AB=1,BC=y,AC=2

By Pythagoras Theorem,

AC

2

=AB

2

+BC

2

2

2

=1

2

+y

2

y

2

=3

y=

3

Now, (secx

0

−tanx

0

)(secx

0

+tanx

0

)

= sec

2

x

∘

−tan

2

x

∘

= (

B

H

)

2

−(

B

P

)

2

=

1

2

−

1

3

= 4−3

= 1

Answer 2

Answer

Open in answr app

Given that,

(i) y=?

(ii) sinx2=?

(iii) (secxo−tanxo)(secxo+tanxo)=?

According to given figure,

By Pythagoras theorem,

(Hypotenuse)2=(perpendicular)2+(Base)2

22=12+y2

4−1=y2

(i) y=3

(ii) sinxo=Hypotenuseperpendicular

sinxo=2y

sinxo=23

(iii) (secxo−tanxo)(secxo+tanxo)

⇒sec2xo−tan2xo

∴(A−

pls mark me as a brainliest....