solution for this

Asked by ABHILASHA | 5th Mar, 2019, 11:18: PM

Expert Answer:

begin mathsize 16px style Given space colon space tanA space plus space cotA space equals space 2
Squaring space on space both space the space sides comma space we space get
open parentheses tanA space plus space cotA close parentheses to the power of 2 space end exponent equals space 2 squared
rightwards double arrow tan squared straight theta space plus space cot squared straight theta space plus space 2 tanθcotθ space equals space 4 space space space space space space space space since space left parenthesis straight a space plus space straight b right parenthesis squared space equals space straight a squared space plus space straight b squared space plus space 2 ab
rightwards double arrow tan squared straight theta space plus space cot squared straight theta space space equals space 4 space minus space 2 tanθcotθ
rightwards double arrow tan squared straight theta space plus space cot squared straight theta space space equals space 4 space minus space 2 space tanθ space cross times space 1 over tanθ
rightwards double arrow tan squared straight theta space plus space cot squared straight theta space equals space 4 space minus space 2

rightwards double arrow tan squared straight theta space plus space cot squared straight theta space equals space space 2 end style

Answered by Yasmeen Khan | 6th Mar, 2019, 10:42: AM