Question

22cosA - 3sinA = 20sinA. Find tan^2A + sin^2A.s3c^2A

Answer 1

given, 22cosA - 3sinA = 20sinA

or, 22cosA = 20sinA + 3sinA

or, 22cosA = 23sinA

or, 22/23 = sinA/cosA = tanA

so, tanA = 22/23 = p/b .......(1)

here, p = 22 and b = 23

from Pythagoras theorem,

h = √(p² + b²) = √(22² + 23²)

= √(484 + 529) = √(1013)

so, sinA = p/h = 22/√(1013) .....(2)

secA = h/b = √(1013)/23 ......(3)

now, tan²A + sin²A. sec²A

from equations (1), (2) and (3),

=(22/23)² + {22/√(1013)}² × {√(1013)/23}

= 484/529 + 484/529

= 968/529

hence, answer should be 968/529

Answer 2

Answer:

Value of the given expression is

$\frac{968}{529}$

Step-by-step explanation:

Given:

$22\:cosA-3\:sinA=20\:sinA$

$22\:cosA=23\:sinA$

$\frac{sinA}{cosA}=\frac{22}{23}$

$tanA=\frac{22}{23}$

Now,

$tan^2A+sin^2A \:sec^2A$

$=tan^2A+sin^2A.\frac{1}{cos^2A}$

$=tan^2A+\frac{sin^2A}{cos^2A}$

$=tan^2A+tan^2A$

$=2tan^2A$

$=2(\frac{22}{23})^2$

$=2(\frac{484}{529})$

$=\frac{968}{529}$