Question

The value of the expression (tan²x + cosec²x + 1) ((1 – sin²x)³ + sin⁶x – 1)) + 3 is

Answer 1

Correct Answer - Option 1 : 0

Formulae Used:

sec x = 1/cos x and cosec x = 1/sin x

sin²x + cos²x = 1

sec²x = 1 + tan²x

(a + b)³ = a³ + b³ + 3ab(a + b)

Calculation:

(sin²x + cos²x)³ = sin⁶x + cos⁶x + 3sin²xcos²x(sin²x + cos²x)

1 = sin⁶x + cos⁶x + 3sin²xcos²x

sin⁶x + cos⁶x – 1 = – 3sin²xcos²x

Now the expression

(tan²x + cosec²x + 1) ((1 – sin²x)³ + sin⁶x – 1)) + 3

⇒ (sec²x + cosec²x)(cos⁶x + sin⁶x - 1) + 3 

⇒ (sec²x + cosec²x)( – 3sin²xcos²x) + 3

⇒ – 3(sec2xsin2xcos2x + cosec2xsin2xcos2x) + 3

⇒ – 3(sin²x + cos²x) + 3

⇒ – 3 + 3

⇒ 0

∴ The value of the expression (tan²x + cosec²x + 1) ((1 – sin²x)³ + sin⁶x – 1)) + 3 = 0

Short Trick: Put x = 45° in the given expression.