(1÷
2Sin A Cos A - SinAX SinA- Cos AX CosA)
Proof this is equal to (1 + SeeAcosecA)
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Step-by-step explanation:
Bro I think your question is wrong from some point.
Because, See this --
LHS =
( 1 ÷ 2sinA cosA - sinA × sinA - cosA × cos A )
= 1 ÷ 2sinAcosA - sin^2A - cos^2A
= 1 ÷ - ( 2sinAcosA + sin^2A - cos^2A )
= -1 ÷ 2sinAcosA + sin^2A - cos^2A
RHS =
= 1 ÷ secA cosecA
= 1 ÷ ( 1 ÷ tanA × 1 ÷ sin A )
= sinA ÷ ( cosA ÷ sinA )
= sin^2A ÷ cosA
-✌️✌️ Please see I may be wrong .
But if you solve this problem please show me the solution .
And if you are preparing for exams , So good luck ☺️☺️☺️
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