Correct Answer - Option 1 : 0

**Formulae Used:**

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

sin^{2}x + cos^{2}x = 1

sec^{2}x = 1 + tan^{2}x

(a + b)^{3} = a^{3} + b^{3} + 3ab(a + b)

**Calculation:**

(sin^{2}x + cos^{2}x)^{3} = sin^{6}x + cos^{6}x + 3sin^{2}xcos^{2}x(sin^{2}x + cos^{2}x)

1 = sin^{6}x + cos^{6}x + 3sin^{2}xcos^{2}x

sin^{6}x + cos^{6}x – 1 = – 3sin^{2}xcos^{2}x

Now the expression

(tan^{2}x + cosec^{2}x + 1) ((1 – sin^{2}x)^{3 }+ sin^{6}x – 1)) + 3

⇒ (sec^{2}x + cosec^{2}x)(cos^{6}x + sin^{6}x - 1) + 3

⇒ (sec^{2}x + cosec^{2}x)( – 3sin^{2}xcos^{2}x) + 3

⇒ – 3(sec2xsin2xcos2x + cosec2xsin2xcos2x) + 3

⇒ – 3(sin^{2}x + cos^{2}x) + 3

⇒ – 3 + 3

⇒ 0

**∴ ****The value of the expression (tan**^{2}x + cosec^{2}x + 1) ((1 **– ****sin**^{2}x)^{3 }+ sin^{6}x **–**** 1)) + 3 = 0**

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