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Rohit

· started a discussion

· 1 Months ago

How (Question:636392.0)

sin²x=cot²x

Question:
Suppose sin⁸ x + 2 sin⁶ x - sin⁴ x – 2 sin² x + 1 = 0 then cot² x + cot⁴ x – 1 = ?
Options:
A) 1 
B) -1
C) -2
D) 0
Solution:
Ans:(d)  sin⁸ x + 2 sin⁶ x - sin⁴ x – 2 sin² x + 1 = 0

(sin⁴ x)² + (sin² x)² + 1² + 2. sin⁴ x. sin² x + 2 sin² x. (-1) + 2(-1) sin⁴ x = 0

(sin⁴ x + sin² x - 1 )² = 0

sin⁴ x + sin² x = 1

Now sin⁴ x = cos² x  and  sin² x = cot² x

We have cot² x + cot⁴ x – 1

= cot² x (1 + cot² x) – 1

= sin² x . cosec² x – 1

= 1 – 1= 0

Knowledge Expert

· commented

· 1 Months ago

Dear student,
Here;
sin^2x * sin^2x = cos^2x
sin^2x = cos^2x/sin^2x = cot^2x

Regards
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