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`        if tan a/2= square root (1+k/1-k)tan b/2 then sec a (cos b - k)/1-kcos b `
one year ago

Arun
24497 Points
```							tana/2=√(1-k/(1+k) tanb/2or, tanb/2=√(1+k)/(1-k) tana/2squaring both sides, tan²b/2=(1+k)/(1-k) tan²a/2or, (1-tan²b/2)/(1+tan²b/2)={(1-k)-(1+k)tan²a/2}/{(1-k)+(1+k)tan²a/2}[by dividendo-componendo method]or, cosb=(1-k-tan²a/2-ktan²a/2)/(1-k+tan²a/2+ktan²a/2)or, cosb{(1-tan²a/2)-k(1+tan²a/2)}/{(1+tan²a/2)-k(1-tan²a/2)}or, cosb=[{(1-tan²a/2)-k(1+tan²a/2)}/(1+tan²a/2)]/                                                   [{(1+tan²a/2)-k(1-tan²a/2)}/(1+tan²a/2)]or, cosb=(cosa-k)/(1-kcosa) [∵, (1-tan²a/2)/(1+tan²a/2)=cosa]   hence on side transwpfer you will get amswer as 1
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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