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प्रश्न
Find sin(x – y), given that sin x = `8/17` with 0 < x < `pi/2`, and cos y = `- 24/25`, x < y < `(3pi)/2`
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उत्तर
sin x = `8/17`, 0 < x < `pi/2`
⇒ Where x is in I quadrant
∴ sin x, cos x are +ve
From ΔABX,
AX = `sqrt(17^2 - 8^2)`
= `sqrt((17 + 8)(17 - 8))`
= `sqrt((25)(9))`
= 5 × 3
= 15
∴ sin x = `8/17` and cos x = `15/17`
cos y = `- 24/25, pi < y < (3pi)/2`
⇒ Where y is in III quadrant
So, sin y and cos y are –ve
From ΔALY,
AL = `sqrt(25^2 - 24^2)`
= `sqrt(49)`
= 7
∴ cos y = `- 24/25` and sin y = `- 7/25`
sin(x – y) = sin x cos y – cos x sin y
= `(8/17)(- 24/25) - (15/17)(- 7/25)`
= `- 192/425 + 105/425`
= `- 87/425`
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