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`        A+B+C=90, ∑ cos(B+C)/cosBcosC   how to do this problem    `
one year ago

1815 Points
```							we have to find out ∑ cos(B+C)/cosBcosC=∑ 1 –  sinBsinC/cosBcosC=∑ 1 – tanBtanC=3 – ∑ tanBtanCnow we have A+B+C=90take tan both sides(tan A + tan B + tan C - tan A tan B tan C)/( 1 - tan A tan B- tan C tan A - tan B tan C)= not defined, which is possible only if 1 - tan A tan B- tan C tan A - tan B tan C=0or ∑ tanBtanC=1so, ∑ cos(B+C)/cosBcosC = 3 – 1= 2
```
one year ago
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Course Features

• 31 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions