Question

pls giys need your help.....
best answer will be marked as brainliest....plz

Answer 1

⇒ tan²α/(tanα - 1) - 1/tanα(tanα -1)

⇒ 1/(tanα - 1) { tan²α - 1/tanα}

⇒ $\frac{tan^3\alpha - 1}{tan\alpha(tan\alpha - 1)}$

⇒ $\frac{(tan\alpha - 1)( 1 + tan\alpha + tan^2\alpha)}{tan\alpha( tan\alpha - 1)}$

⇒$\frac{ 1 + tan\alpha + tan^2\alpha}{tan\alpha}$

⇒ 1 + tanα + cotα

• the given expression was tanα/(1 - cotα) + cotα/(1 - tanα)
• i substituted cotα = 1/tanα
• then the question becomes tan²α/(tanα - 1) + 1/tanα(1 - tanα)
• in order to take 1/(tanα - 1) common i multiplied the numerator and denominator of the second term with -1 so that 1/tanα(1 - tanα) becomes -1/tanα(tanα - 1)
• now the question becomes tan²α/(tanα - 1) - 1/tanα(tanα -1)
• i wrote the above expression directly by reading this summary you can understand the problem without any doubt. . thank you.
• a³ - b³ = (a - b)( a² + ab + b²)