Question

how to solve sec A +tanA=3=secA​

Answer 1

Answer:

1

Step-by-step explanation:

Given,

         α + β = 180( in degree )

⇒ α = 180 - β        

⇒ sinα = sin( 180 - β )

⇒ sinα = sinβ           { using sin( π - B ) = sinB }

⇒ sin^2 α = sin^2 β            ....( 1 )

Therefore,

⇒ sin^2 α + cos^2 β

⇒ sin^2 β + cos^2 β          { from ( 1 ) }

⇒ 1                      { we know, sin^2 A + cos^2 A = 1 }

Answer 2

Step-by-step explanation:

GIVEN:

$\alpha and \beta$ are supplementary.

I. e. $\alpha +\beta=180°$

TO FIND:

$sin^{2}\alpha + cos^{2}\beta$

SOLUTION:

=>$\alpha +\beta=180°$

=>$\alpha=180°-\beta$

On using sin both sides,

=>$sin\alpha=sin(180°-\beta)$

$\large\green{\boxed{sin(180°-\theta)=sin\theta) }}$

=>$sin\alpha=sin\beta$

On squaring both sides,

=>$sin^{2}\alpha=sin^{2}\beta$

...............(1)

________________________________

Now,

=$sin^{2}\alpha + cos^{2}\beta$

=$sin^{2}\beta+cos^{2}\beta$

(using equation1)

=$\large\red{\boxed{sin^{2}\theta+cos^{2}\theta=1}}$

=$1$

$\huge\orange{\boxed{ANSWER:1}}$