sin squar A. cos squar B+cos squar A.sin squar B+cos squar A. cos squar B+ sin squar A. sin squar B=1 prove that
Answers
Pythagoras in Disguise
This formula is the Pythagorean theorem in disguise.
{\displaystyle \sin ^{2}(\theta )+\cos ^{2}(\theta )=1} {\displaystyle \sin ^{2}(\theta )+\cos ^{2}(\theta )=1}
If you look at the diagram in the next section it should be clear why. Sine and Cosine of an angle {\displaystyle \theta } \theta in a triangle with unit hypotenuse are just the lengths of the two shorter sides. So squaring them and adding gives the hypotenuse squared, which is one squared, which is one.
Pythagorean Trigonometric Identity
Similar right triangles showing sine and cosine of angle θ
In more detail... In a right triangle with sides {\displaystyle a,b} a,b and hypotenuse {\displaystyle c} c, trigonometry (Soh-Cah-Toa) determines the sine and cosine of the angle {\displaystyle \theta } \theta between side {\displaystyle a} a and the hypotenuse as:
{\displaystyle \sin(\theta )={\frac {b}{c}}\,\ \quad \cos(\theta )={\frac {a}{c}}} {\displaystyle \sin(\theta )={\frac {b}{c}}\,\ \quad \cos(\theta )={\frac {a}{c}}}
From that it follows:
{\displaystyle \sin ^{2}(\theta )+\cos ^{2}(\theta )={\frac {b^{2}+a^{2}}{c^{2}}}=1} {\displaystyle \sin ^{2}(\theta )+\cos ^{2}(\theta )={\frac {b^{2}+a^{2}}{c^{2}}}=1}
where the last step applies Pythagoras' theorem. This relation between sine and cosine sometimes is called the fundamental Pythagorean trigonometric identity.[1] In similar triangles, the ratios of the sides are the same regardless of the size of the triangles, and depend upon the angles. Consequently, in the figure, the triangle with hypotenuse of unit size has opposite side of size {\displaystyle \sin(\theta )} {\displaystyle \sin(\theta )} and adjacent side of size {\displaystyle \cos(\theta )} {\displaystyle \cos(\theta )} in units of the hypotenuse.
Neither sine nor cosine can ever exceed 1 and the closer one of them is to 1, the closer the other must be to 0. We can see this in two ways:
It follows immediately from the formula. As either sine squared or cosine squared gets closer to one the amount left for the other diminishes.
It can be seen from the geometry. The hypotenuse is one and is longer than either of the other sides. As one side gets closer to one, the other must get closer to 0
As we know tan A=Sin A/Cos A and tan B=Sin B/Cos B So
tan squareA - tan squareB= (Sin A/Cos A)^2 - (Sin B/Cos B)^2 =
{(Sin^2 A Cos^2 B)-(Cos^2 A Sin^2 B)}/(Cos^2 A .Cos^2 B)
As we know that Sin^2 A=1-Cos^2 A and Sin^2 B=1-Cos^2 B ,So
tan squareA - tan squareB={( Cos^2 B(1-cos^2 A))-(Cos^2 A (1-Cos^2 B))}/(Cos^2 A .Cos^2 B)= (Cos^2 B- Cos^2 A)/(Cos^2 A .Cos^2 B)
Read more on Brainly.in - https://brainly.in/question/732956#readmore