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(1-cosa)(1-cosb)(1-cosc)=sinasinbsinc find the value of (1=cosa)(1=cosb)(i+cosc)

Abhijeet Sahdev , 9 Years ago
Grade 11
anser 7 Answers
Lab Bhattacharjee

Last Activity: 9 Years ago

\text{HINT: }\\ (1+\cos a)(1-\cos a)(1+\cos b)(1-\cos b)(1+\cos c)(1-\cos c)=\sin^2a\sin^2b\sin^2c

Lab Bhattacharjee

Last Activity: 9 Years ago

\text{HINT: }\\ (1+\cos a)(1-\cos a)(1+\cos b)(1-\cos b)(1+\cos c)(1-\cos c)=\sin^2a\sin^2b\sin^2c

Lab Bhattacharjee

Last Activity: 9 Years ago

\text{HINT: }\\ (1+\cos a)(1-\cos a)(1+\cos b)(1-\cos b)(1+\cos c)(1-\cos c)=\sin^2a\sin^2b\sin^2c

Lab Bhattacharjee

Last Activity: 9 Years ago

\text{HINT: }\\ (1+\cos a)(1-\cos a)(1+\cos b)(1-\cos b)(1+\cos c)(1-\cos c)=\sin^2a\sin^2b\sin^2c

Lab Bhattacharjee

Last Activity: 9 Years ago

\text{HINT: }\\ (1+\cos a)(1-\cos a)(1+\cos b)(1-\cos b)(1+\cos c)(1-\cos c)=\sin^2a\sin^2b\sin^2c

Lab Bhattacharjee

Last Activity: 9 Years ago

To the administrator, please delete the repeated answer(s).
 

Raghu Vamshi Hemadri

Last Activity: 9 Years ago

\text{HINT: }\\ (1+\cos a)(1-\cos a)(1+\cos b)(1-\cos b)(1+\cos c)(1-\cos c)=\sin^2a\sin^2b\sin^2c

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