Given SiNx + cosx = (root 3 + 1) / 2, find the value of SiNx / (1-1 / TaNx) + cosx / (1-tanx) After thinking for a long time, I still have no idea. Please,

Given SiNx + cosx = (root 3 + 1) / 2, find the value of SiNx / (1-1 / TaNx) + cosx / (1-tanx) After thinking for a long time, I still have no idea. Please,

sinx/（1-1/tanx）+cosx/(1-tanx)=sinxtanx/(tanx-1)-cosx/(tanx-1)
=Cosx [(TaNx) ^ 2-1] / (tanx-1) = cos (TaNx + 1) = SiNx + cosx = (root 3 + 1) / 2

Given SiNx + cosx = - root sign 10 / 5, find the value of (1) 1 / SiNx + 1 / cosx (2) TaNx Given SiNx + cosx = - root sign 10 / 5, find (1) Value of 1 / SiNx + 1 / cosx (2) TaNx

Because SiNx + cosx = - (root 10) / 5, so (SiNx + cosx) ^ 2 = [- (root 10) / 5] ^ 2, so (SiNx) ^ 2 + 2 * SiNx * cosx + (cosx) ^ 2 = 2 / 5, because (SiNx) ^ 2 + (cosx) ^ 2 = 1, so 1 + 2 * SiNx * cosx = 2 / 52 * SiNx * cosx = - 3 / 5sinx * cosx = - 3 / 10, so 1 / SiNx + 1 / cosx = (cosx + SiNx) /

We already know the value of cosx and how to find SiNx and TaNx in the interval of X

SiNx square = 1-cosx square (see whether to add a minus sign according to the square of the interval), TaNx = SiNx / cosx

If TaNx is = 2, the value of 2sinx cosx / SiNx + cosx

（2sinx-cosx）/（sinx+cosx）
=(2tanx-1) / (TaNx + 1) (numerator, denominator, same as division by cosx)
=(2*2-1)/(2+1）
=3/3
=1

If (1-sinx / 1 + SiNx) = tanx-1 / cosx under the root sign, find the range of angle X

Under the root sign (1-sinx / 1 + SiNx) = |cosx| / (1 + SiNx) = (1-sinx) / |cosx|
tanx－1/cosx＝-(1-sinx)/cosx
Therefore, cosx < 0, the range of X is 2K π + π / 2 < x < 2K π + 3 π / 2, and K is any integer

A point P (- root sign 3, m) on the final edge of angle X, and SiNx = (root sign 2) / m, find the values of cosx and TaNx

The writing is very complicated. Only the solution is provided:
SiNx ^ 2 + cosx ^ 2 = 1, cosx can be obtained. Note: the quadrant of X may make the result different from the sign
From a point P (- radical 3, m) on the final edge of angle X, TaNx = - M / radical 3 is obtained,