Question

tan∅/1-cot∅ + cot∅/1-tan∅ = 1+ sec∅ cosec∅
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Answer 1

Step-by-step explanation:

\begin{gathered}\frac{tanx}{1-cotx}+\frac{cotx}{1-tanx}\\\\=\frac{tanx}{\frac{tanx-1}{tanx}}+\frac{1}{tanx}\frac{1}{1-tanx}\\\\=\frac{tan^2x}{tanx-1}-\frac{1}{tanx\ (1- tanx)}\\\\=\frac{tan^3x-1}{tanx(tanx-1)}\\\\=\frac{(tanx-1)(tan^2x+tanx+1)}{tanx(tanx-1)}\\\\if\ tanx\ \neq\ 1,\ then:\\\\=tanx+1+Cotx\\\\=1+\frac{sinx}{cosx}+\frac{cosx}{sinx}\\\\=1+\frac{sin^2+cos^2x}{sinx\ cosx}\\\\=1+secx\ cosecx\end{gathered}

1−cotx

tanx

+

1−tanx

cotx

=

tanx

tanx−1

tanx

+

tanx

1

1−tanx

1

=

tanx−1

tan

2

x

−

tanx (1−tanx)

1

=

tanx(tanx−1)

tan

3

x−1

=

tanx(tanx−1)

(tanx−1)(tan

2

x+tanx+1)

if tanx



= 1, then:

=tanx+1+Cotx

=1+

cosx

sinx

+

sinx

cosx

=1+

sinx cosx

sin

2

+cos

2

x

=1+secx cosecx