Math, asked by sonali3971, 1 year ago

plzzz give me answer .............

Attachments:

Answers

Answered by mrhappy1987
1

LHS=

1−cosθ

tanθ

+

1−tanθ

cotθ

=

1−cotθ

tanθ

+

1−

cotθ

1

cotθ

\begin{lgathered}=\frac{tan{\theta}}{1-cot{\theta}}-\frac{cot^2{\theta}}{1-cot{\theta}} \\\end{lgathered}

=

1−cotθ

tanθ

1−cotθ

cot

2

θ

\begin{lgathered}=\frac{\frac{1}{cot{\theta}}-cot^2{\theta}}{1-cot{\theta}} \\\end{lgathered}

=

1−cotθ

cotθ

1

−cot

2

θ

\begin{lgathered}=\frac{1-cot^3{\theta}}{cot{\theta}(1-cot{\theta})} \\\end{lgathered}

=

cotθ(1−cotθ)

1−cot

3

θ

\begin{lgathered}=\frac{(1-cot{\theta})(1+cot^2{ \theta}+cot{\theta})}{cot{\theta}(1-cot{\theta})} \\\end{lgathered}

=

cotθ(1−cotθ)

(1−cotθ)(1+cot

2

θ+cotθ)

\begin{lgathered}=\frac{1}{cot{\theta}}+\frac{cot^2{\theta}}{cot{\theta}}+\frac{cot{\theta}}{cot{\theta}} \\\end{lgathered}

=

cotθ

1

+

cotθ

cot

2

θ

+

cotθ

cotθ

= tan{\theta}+cot{\theta}+1=RHS=tanθ+cotθ+1=RHS

Similar questions