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If cos α = a and sin α = b, then the value of sin²β in terms of a and b is cos β sin β
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a² + 1 a² - b² -
a² - b² a² + b² -
a² - 1 a² - b² -
a² - 1 a² + b²
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Correct Option: C
| = a ⇒ cosα = a cosβ | cos β |
On squaring both sides,
cos²α = a² cos²β
⇒ 1 – sin²α = a² (1 – sin²β) ...(i)
Again, sinα = b sinβ
⇒ sin²α = b² sin²β
∴ From equation (i),
1 – b² sin²β = a² – a² sin²β
⇒ a² sin²β – b² sin²β = a² – 1
⇒ sin²β (a² – b² ) = a² – 1
| ⇒ sin²β = | a² - b² |
