Question

1 by sec theta minus tan theta minus 1 by cos theta is equal to 1 by cos theta minus 1 by sec theta + tan theta​

Answer 1

Answer:

Given that, we have to prove

\frac{1}{sec\ \theta - tan\ \theta} - \frac{1}{cos\ \theta} = \frac{1}{cos\ \theta} - \frac{1}{sec\ \theta + tan\ \theta}

sec θ−tan θ

1

−

cos θ

1

=

cos θ

1

−

sec θ+tan θ

1

Take the LHS:

\begin{gathered}\frac{1}{sec\ \theta - tan\ \theta} - \frac{1}{cos\ \theta}\\\\\frac{1}{sec\ \theta - tan\ \theta} \times \frac{sec\ \theta + tan\ \theta}{sec\ \theta + tan\ \theta} - \frac{1}{cos\ \theta}\\\\\frac{sec\ \theta + tan\ \theta}{sec^2 \theta - tan^2 \theta} - \frac{1}{cos\ \theta}\\\\\end{gathered}

sec θ−tan θ

1

−

cos θ

1

sec θ−tan θ

1

×

sec θ+tan θ

sec θ+tan θ

−

cos θ

1

sec

2

θ−tan

2

θ

sec θ+tan θ

−

cos θ

1

We know that,

sec^2\ \theta - cos^2\ \theta = 1sec

2

θ−cos

2

θ=1

Also,

sec\ \theta = \frac{1}{cos\ \theta}sec θ=

cos θ

1

Therefore,

\begin{gathered}\frac{sec\ \theta + tan\ \theta}{1} - sec\ \theta\\\\sec\ \theta + tan\ \theta - sec\ \theta\\\\tan\ \theta\end{gathered}

1

sec θ+tan θ

−sec θ

sec θ+tan θ−sec θ

tan θ

Now take the RHS:

\begin{gathered}\frac{1}{cos\ \theta} - \frac{1}{sec\ \theta + tan\ \theta}\\\\ \frac{1}{cos\ \theta} - \frac{1}{sec\ \theta + tan\ \theta} \times \frac{sec\ \theta - tan\ \theta}{sec\ \theta - tan\ \theta}\\\\ \frac{1}{cos\ \theta} - \frac{sec\ \theta - tan\ \theta}{sec^2 \theta -tan^2 \theta}\\\\ \frac{1}{cos\ \theta} - \frac{sec\ \theta - tan\ \theta}{1}\\\\sec\ \theta - sec\ \theta + tan\ \theta\\\\tan\ \theta\end{gathered}

cos θ

1

−

sec θ+tan θ

1

cos θ

1

−

sec θ+tan θ

1

×

sec θ−tan θ

sec θ−tan θ

cos θ

1

−

sec

2

θ−tan

2

θ

sec θ−tan θ

cos θ

1

−

1

sec θ−tan θ

sec θ−sec θ+tan θ

tan θ

Thus,

LHS = RHS

Thus proved

Explanation:

Hope it helps

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