) S AT P ;1 —sin30° ८०560 + ८०5301+ 5in?45 290°

1.	) S AT P ;1 —sin30° ८०560 + ८०5301+ 5in?45 290° — cot290°= + (sin60"tan30")
Answer» sin30=1/2sin45=1/√2cos60=1/2cos30=√3/2cosec90=1cot90=0sin60=√3/2tan30=1/√3Hence put in the equation1-1/4/1+1/2(1/4+3/4/1-0)÷(√3/21/√3)=3/4/3/2(1/1)÷(1/2)=2÷1/2=22=4 Thanks..I am justing testing this app

) S AT P ;1 —sin30° ८०560 + ८०5301+ 5in?45 290° — cot290°= + (sin60"tan30")

Answer»

sin30=1/2sin45=1/√2cos60=1/2cos30=√3/2cosec90=1cot90=0sin60=√3/2tan30=1/√3Hence put in the equation1-1/4/1+1/2*(1/4+3/4/1-0)÷(√3/2*1/√3)=3/4/3/2*(1/1)÷(1/2)=2÷1/2=2*2=4

Thanks..I am justing testing this app

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) S AT P ;1 —sin30° ८०560 + ८०5301+ 5in?45 290° — cot290°= + (sin60"tan30")

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