SecA(1-sinA) (secA + tanA) =1
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Answer:
Step-by-step explanation:
SecA(1-SinA).(SecA+TanA)=1
LHS
1/CosA(1-SinA) . (1/CosA+SinA/CosA).... [Sec=1/Cos and Tan=Sin/Cos]
(1-SinA/Cos A) . (1+SinA/CosA)
(1-SinA)(1+SinA)/Cos²A
1²-Sin²A/Cos²A......... [(a+b).(a-b)=a²-b²]
1-Sin²A/Cos²A
Cos²A/Cos²A..... [sin²A+cos²A=1⇒cos²A=1-sin²A]
= 1
Hence, PROVED
LHS=RHS
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