Question

answer me 1st questions plzz

Answer 1

Hey there !!!

cosα/sin³α+cos³α=???

tanα=sinα/cosα=2

tanα=opposite side/adjacent side =2/1

So, opposite side =2 adjacent side =1

hypotenuse²=opposite side²+adjacent side²

x²=2²+1²

x²=5

x=+√5 or -√5

As α llies in between ( π,3π/2) tanα is positive because α lies in third quadrant and tanα is positive and sinα and cosα are negative in third quadrant.

sinα=opposite/hypotenuse = -2/√5 

cosα=adjacent side/hypotenuse =-1/√5

cosα/sin³α+cos³α = (-1/√5)/( (-2/√5)³+( -1/√5)³)

                             =-1/√5/-8/5√5-1/5√5

                              =-1/√5/-9/5√5
                              
   cosα/sin³α+cos³α =5/9


Hope this helped you..........

Answer 2

Hello Friend,

Here's the solution.
You need to realise that α lies in Third Quadrant.
In third quadrant, sine and cosine functions are negative.


Hope it helps.