Question

pls answer fast!!
WILL MARK THE BRAINLIEST ASAP!! ​

Answer 1

Step-by-step explanation:

w. k. t,,

sec^2-tan^2=1

sec+ tan(sec-tan) =1

sec+ tan=1/(sec-tan)......... eq1

w. k. t,,

sin theta=a/b

therefore opp side of angle theta=a

hypotenuse side of angle theta=b

by pythogoras thm adj side of angle theta theta=√b^2-a^2

continuation of eq1,,

sec+ tan=1/[(1-sin)/cos]

sec+ tan=cos/1-sin

substitute,,

cos theta=(√b^2-a^2)/b

sin theta=a/b

sec+ tan=(√b^2-a^2)/b÷1-(a/b)

we get (√b-a*√b+a)/b-a

b-a can be written as √b-a*√b-a

therfore, sec+ tan=√b+a/√b-a

sec+ tan=√(b+a/b-a)