Question

show that √(sec^2 theta + cosec^2 theta) = tan theta + cot theta​

Answer 1

Answer:

√(sec^2 + Cosec^2)

√( 1/cos^2 + 1/sin^2 )

take l.cm as sin^2cos^2

so we

√(sin^2 + cos^2/sin^2cos^2)

as Sin^2 +cos^2 = 1

√ (1/sin^2cos^2 )

√(1/sin theta cos theta )^2

square and root cancel

we 1/sintheta cos theta

similarly

in r.h.s

# tan +cot

# sin/cos +cos/sin

# taken l.cm sincos

we get

sin^2 +cos^2 / sincos

1/sincos

hence prove