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MODERN PUBLICATION-TRIGONOMETRY-NCERT(EXERCISE-3.3)
- sin^2(pi/6)+cos^2(pi/3)-tan^2(pi/4)=-1/2
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- Prove that: 2s in^2pi/6+cos e c^2(7pi)/6cos^2pi/3=3/2
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- cot^(2)pi/6+"cosec"(5pi)/(6) + 3tan^(2)pi/6=6
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- 2sin^2(3pi)/4+2cos^2pi/4+2sec^2pi/3=10
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- Find the value of: "sin"75^(@) "tan"15^(@)
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- Prove the following cos(pi/4-x)cos(pi/4-y)-sin(pi/4-x)sin(pi/4-y)=sin...
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- (tan(pi/4+x))/(tan(pi/4-x))=((1+tanx)/(1-tanx))^2
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- (cos(pi+x)cos(-x))/(sin(pi-x)cos(pi/2-x))=cot^2x
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- cos((3pi)/(2) +x) cos(2pi+x)[cot(3pi)/(2)-x+cot(2pi+x)]=1
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- sin(n+1)xsin(n+2)x+cos(n+1)xcos(n+2)x=cosx
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- Prove that: cos((3pi)/(4)+A)-cos((3pi)/(4)-A)=-sqrt(2)sinA
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- sin^(2)6x-sin^(2)4x=sin2xsin10x
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- Prove the following cos^(2)2x-cos^(2)6x="sin"4x"sin"8x
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- Prove the following "sin"2x+2"sin"4x+"sin"6x=4cos^(2)x"sin"4x
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- Prove that: cot" "4x" "(s in" "5x" "+" "s in" "3x)" "=" "cot" "x" "(s ...
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- (cos 9x - cos 5x)/(sin 17x - sin 3x) = (-sin 2x)/(cos 10x)
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- Prove the following ("sin"5x+"sin"3x)/(cos5x-cos3x)="tan"4x
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- Prove the following ("sin"x-"sin"y)/(cosx+cosy)="tan""(x-y)/(2)
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- Prove the following ("sin"x+"sin"3x)/(cosx-cos3x)="tan"2x
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- Prove the following ("sin"x-"sin"3x)/("sin"^(2)x-cos3x)=2"sin"x
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