Hello aspirant
Today we are going to learn all inverse trigonometry function but before this you need to learn the trigonometry formulas make sure if didn't go through that so it make difficulty for you.
So let start with all your Josh !!!!!
- Sin(sin−1x)=x
- Tan(tan−1x)=x
- Sec(sec−1x)=x
- Cot(cot−1x)=x
- Cos(cos−1x)=x
- Cosec (cosec−1x)=x
>sin−1 (-x)= -sin−1x
>cos−1(-x)=Ï€ -cos−1x
>cosec−1(-x)= -cosec−1x
>sec−1(-x)=Ï€ -sec−1x
>tan−1(-x)= -tan−1x
>cot−1 (-x)= Ï€-cot−1x
>sin−1x + cos−1x =Ï€/2
>tan−1x + cot−1x=Ï€/2
>sec−1x + cosec−1x=Ï€/2
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>2tan−1x=sin−1(2x/1+x²)=cos−1(1-x²/1+x²)
>2tan−1x=tan−1(2x/1-x²)
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>tan−1x=sin−1 (x/√1+x²)=cos−1x(1/√1+x²)
>sin−1x =tan−1x (x/√1-x²)
>cos−1x =tan−1(√1-x²/x)
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>sin−1x=cos−1√1-x²
>cos−1x=sin−1√1-x²
>Sin(cos−1x)=cos(sin−1x)=√1-x²
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>sin−1x+sin−1y=x√1-y² + y√1-x²
>sin−1x-sin−1y=x√1-y² - y√1-x²
>sin−1x =tan−1x (x/√1-x²)
>cos−1x =tan−1(√1-x²/x)
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>sin−1x=cos−1√1-x²
>cos−1x=sin−1√1-x²
>Sin(cos−1x)=cos(sin−1x)=√1-x²
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>sin−1x+sin−1y=x√1-y² + y√1-x²
>sin−1x-sin−1y=x√1-y² - y√1-x²
>cos−1x +cos−1y= xy–√1-x² √1-y²
>cos−1x –cos−1y= xy+√1-x² √1-y²
>tan−1x +tan−1y =tan−1(x+y/1–xy) xy<1
>tan−1x –tan−1y =tan−1(x–y/1+xy)
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>3sin−1x=sin−1(3x–4x³)
>3cos−1x=cos−1x(4x³–3x)
>3tan−1x =tan−1 (3x–x³/1–3x²)
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>2sin−1x=sin−1 2x √1–x²
>2cos−1x=sin−1 2x √1–x² =cos−1(2x²–1)
>tan−1x +tan−1y =tan−1(x+y/1–xy) xy<1
>tan−1x –tan−1y =tan−1(x–y/1+xy)
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>3sin−1x=sin−1(3x–4x³)
>3cos−1x=cos−1x(4x³–3x)
>3tan−1x =tan−1 (3x–x³/1–3x²)
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>2sin−1x=sin−1 2x √1–x²
>2cos−1x=sin−1 2x √1–x² =cos−1(2x²–1)
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