Question

A+B=90 then simplest form root sinAsecB - sinAcosB

Answer 1

Answer:

cos A

Step-by-step explanation:

Given sinA*secB_sinA*cosB

secB=1/cosB

=[sinA/cosB]_[sinA*cosB]

cos B=cos(90_A) = sinA

= sinA/sin A _ sinA*sin A

=1 _(sinA)^

BY THE IDENTITY ( cosA)^2 + (sinA)^2 = 1

then =(cosA)^2

Answer 2

Answer:

$\sqrt{sin \: A \: sec B - sin \: A \: cos \: B} = cos \: A$

Step-by-step explanation:

If

A+B =90°

To find the simplest form of

$\sqrt{sin \: A \: sec \: B - sin \: A \: cos \: B}$

As we know that

$sec(90 - \theta) = cosec \: \theta \\cos(90 - \theta) = sin \: \theta$

So put the value of b in terms of a

$\sqrt{sin \: A \: sec \: (90 - A) - sin \: A \: cos \: (90 -A)}$

$\sqrt{sin \: A \: cosec \: A- sin \: A \: sin \: A}$

$\sqrt{sin \: A \times\:\frac{1}{sin \: A} - {sin}^{2} \: A \: }$

$=\sqrt{1 - {sin}^{2}A }$

$= \sqrt{ {cos}^{2}A }$

$= cos \: A$

$\because\:1 - {sin}^{2}A={cos}^{2}A }$

Hope it helps you.