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tan1. prove that sec 0tan30 cosec 9222 |
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Answer» Tan²θ=1-a²LHS=secθ+tan³θcosecθ=√(1+tan²θ)+tan²θ×tanθ×√(1+cot²θ)[∵, sec²θ-tan²θ=1 and cosec²θ-cot²θ=1]=√1+(1-a²)+(1-a²)×√(1-a²)×√{1+(1/tan²θ)}=√(2-a²)+(1-a²)×√(1-a²)×√{1+1/(1-a²)}=√(2-a²)+(1-a²)×√(1-a²)×√{(1-a²+1)/(1-a²)}=√(2-a²)+(1-a²)×√(2-a²)=√(2-a²)×(1+1-a²)=√(2-a²)×(2-a²)=(2-a²)¹/²⁺¹=(2-a²)³/²=RHS (Proved) |
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