Prove that: \$sin ,cot^-1 cos , an^-1x = sqrtdfracx^2+1x^2+2 \$.

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This question was taken from Miscellaneous Example of S.L. Loney"s Trigonometry.

Since, the question involves 2 parts, LHS và RHS, LHS is completely Trigonometric while, RHS is algebraic, so my approach was to get taylor series for \$x+1\$ then differentiate it và keep on doing to lớn show that both sides are same... But, I wonder if there is an trigonometric solution for the question too or not!

When you compose trig functions with inverse trig you get algebraic functions. I always draw a right triangle. If we first assume the angle is in the first quadrant, we have
where \$ an^-1x\$ is angle \$A\$ & \$cos A=frac 1sqrt1+x^2\$. Now draw a new triangle for the next two functions. Then you need to lớn make sure it works for other quadrants because of the ranges of the inverse trig functions.

Let \$y= an^-1x\$. Then \$ an y=x\$ and so drawing a right triangle with angle \$y\$ and sides \$1,x,sqrt1+x^2\$, we find \$cos y=dfrac1sqrt1+x^2\$.

Now let \$t=cot^-1(cos y)\$. Then \$cot t=cos y=dfrac1sqrt1+x^2\$, so drawing a right triangle with angle \$t\$, we find the hypotenuse is \$sqrt2+x^2\$, so \$sin t=sqrtdfrac1+x^22+x^2\$.

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