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

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

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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.

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