Question

If A + B = 90, prove that √tana tanB + tana cotB/sin A secB= secA.

Answer 1

HELLO DEAR,

GIVEN:- A + B = 90°
A = 90° - B----------( 1 )

now, $\sf{\sqrt{(tanAtanB + tanAcotB)}/sinAsecB}$

$\sf{\sqrt{[tan(90 - B)tanB + tan(90 - B)cotB]}/sin(90 - B) secB}$

$\sf{\sqrt{(cotBtanB + cotBcotB)}/cosB*1/cosB}$

$\sf{\sqrt{(1 + cot^2B)}/1}$

$\sf{\sqrt{cosec^2B}}$

$\sf{cosec(90 - A)}$

$\sf{= sec A}$

HENCE, L.H.S = R.H.S

I HOPE ITS HELP YOU DEAR,
THANKS

Answer 2

Hi ,
**************************************
It is given that ,

A + B = 90°

A = 90 - B

tan A = tan ( 90 - B )

= cotB ---( 1 )

SinA = sin ( 90 - B ))

= cos B ---( 2 )

sec B = 1/cosB ---( 3 )

1 + cot² B = cosec² B

*******************************************
LHS = √[(tanAtanB+tanAtanB)/(sinAsecB)]

= √[(cotBtanB+cotBcotB )/(cosB×1/cosB)]

= √[(1 + cot²B )

=√cosec²B

= cosec B

= cosec ( 90 - A )

= sec A

= RHS

I hope this helps you.

: )