# If p.tan(a)=tan(p.a) then show that, sin^2(pa)/sin^2(a)=p^2/{1+(p^2-1).sin^2(a)}please tell the detail procedure,and describe how you are going upper step from the lower

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P*(a)= tan(pa)

We know that sin(a)= tan(a)/(1+tan^2(a))^1/2

==> sin^2(a)= tan^2(a)/[1+tan^2(a)]......(1)

==> sin(pa) = tan(pa)/(1+tan^2(pa))^1/2

= ptan(a)/[1+(p^2)*tan^2(a)]^1/2

==> sin^2(pa) = p^2tan^2(a)/(1+(p^2)tan^2(a)]...(2)

Divide (2) by (1)...

==> sin^2(pa)/sin^2(a) =[p^2*tan^2(a)]*[1+tan^2(a)]/(tan^2(a)*(1+P^2*tan^2(a)

==> after simplifying:

sin^2(pa)/sin^2(a)= p^2/{1+(p^2-1)sin^2(a)}

If p*tana = tan(p.a). To show that sin^2(p.a)/(sin^a) = {1+(p^2-1) sin^2(a)}

Solution:

We know that if tan x = t, then sinx = t/sqrt(1+t^2). Using this idea in the current problem, we get:

tan(p.a) = p.tana , given. Therefore using the above,

sin (p.a) = (ptana)/sqrt[1+(ptana)^2]. Or squaring both sides, we get:

sin^2 (pa) = (p^2 tan^2a/(1+ p^2tan^2a)

= p^2*sin^2a / (cos^2a + p^2sin^2a). Or dividing by sin^2a both sides, we get:

sin^2(pa)/sin^2a = p^2/{ 1-sin^2 a+p^2sin^2a). simplifying or rearranging the right side after replacing cos^2 by 1-sin^2a , we get:

= p^2 / [1+ (p^2-1)sin^2a]. Or

sin^2(pa) /sin^2a = p^2/{1+(p^2-1) sin^2 a}

Note: tan(pa) = p*tana holds only for when p and a are small (around zero).