Question

12 If 22 cos A - 3 sin A = 20 sin A, then find the value of
tan’ A+sin? A sec? A.​

Answer 1

given,

22 cosA - 3 sinA = 20 sinA

or, 22 cosA = 20 sinA + 3 sinA

or, 22 cosA = 23 sinA

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}

=(22/23)² + {22/√(1013)}² × {√(1013)/23}= 484/529 + 484/529

=(22/23)² + {22/√(1013)}² × {√(1013)/23}= 484/529 + 484/529= 968/529

hence, answer should be 968/529

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