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Question : Given B + C = πᶜ. Prove that, 2 (1 - sinB sinC) = cos²C + cos²B.
Trigonometry : Trigonometry is the study of angles and relations between angles and their sin, cos, tan, cosec, sec, cot ratios. There are many formulae for calculations:
• sin²A + cos²A = 1
• sec²A - tan²A = 1
• cosec²A - cot²A = 1
• sinA + sinB = 2 sin{(A + B)/2} cos{(A - B)/2}
• sinA - sinB = 2 sin{(A - B)/2} cos{(A + B)/2}
• cos(π/2) = 0
Proof :
Given, B + C = πᶜ
Now, sinB - sinC
= 2 sin{(B - C)/2} cos{(B + C)/2}
= 2 sin{(B - C)/2} cos(π/2)
= 2 sin{(B - C)/2} × 0
= 0
∴ R.H.S. = cos²C + cos²B
= 1 - sin²C + 1 - sin²B
= 2 - (sin²B + sin²C)
= 2 - {(sinB - sinC)² + 2 sinB sinC}
= 2 - {0² + 2 sinB sinC}
= 2 - 2 sinB sinC
= 2 (1 - sinB sinC) = L.H.S.
Hence, proved.
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