If θ1,θ2,θ3,…,θn are in AP, whose common difference is d, then  sin⁡ dsec⁡θ1sec⁡θ2+sec⁡θ2sec⁡θ3+…+sec⁡θn−1sec⁡θn is equal to

# If ${\mathrm{\theta }}_{1},{\mathrm{\theta }}_{2},{\mathrm{\theta }}_{3},\dots ,{\mathrm{\theta }}_{\mathrm{n}}$ are in AP, whose common difference is d, then $+\dots +\mathrm{sec}{\mathrm{\theta }}_{\mathrm{n}-1}\mathrm{sec}{\mathrm{\theta }}_{\mathrm{n}}\right)$ is equal to

1. A

$\mathrm{tan}{\mathrm{\theta }}_{\mathrm{n}}-\mathrm{tan}{\mathrm{\theta }}_{2}$

2. B

$\mathrm{tan}{\mathrm{\theta }}_{\mathrm{n}}+\mathrm{tan}{\mathrm{\theta }}_{1}$

3. C

$\mathrm{tan}{\mathrm{\theta }}_{\mathrm{n}}-\mathrm{tan}{\mathrm{\theta }}_{1}$

4. D

None of these

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### Solution:

Since,${\mathrm{\theta }}_{1},{\mathrm{\theta }}_{2},{\mathrm{\theta }}_{3},\dots ,{\mathrm{\theta }}_{\mathrm{n}}$ are in AP

Now, taking only first term

$\begin{array}{l}\mathrm{sin}\mathrm{dsec}{\mathrm{\theta }}_{1}\mathrm{sec}{\mathrm{\theta }}_{2}=\frac{\mathrm{sin}\mathrm{d}}{\mathrm{cos}{\mathrm{\theta }}_{1}\mathrm{cos}{\mathrm{\theta }}_{2}}=\frac{\mathrm{sin}\left({\mathrm{\theta }}_{2}-{\mathrm{\theta }}_{1}\right)}{\mathrm{cos}{\mathrm{\theta }}_{1}\mathrm{cos}{\mathrm{\theta }}_{2}}\\ =\frac{\mathrm{sin}{\mathrm{\theta }}_{2}\mathrm{cos}{\mathrm{\theta }}_{1}-\mathrm{cos}{\mathrm{\theta }}_{2}\mathrm{sin}{\mathrm{\theta }}_{1}}{\mathrm{cos}{\mathrm{\theta }}_{1}\mathrm{cos}{\mathrm{\theta }}_{2}}\\ =\frac{\mathrm{sin}{\mathrm{\theta }}_{2}\mathrm{cos}{\mathrm{\theta }}_{1}}{\mathrm{cos}{\mathrm{\theta }}_{1}\mathrm{cos}{\mathrm{\theta }}_{2}}-\frac{\mathrm{cos}{\mathrm{\theta }}_{2}\mathrm{sin}{\mathrm{\theta }}_{1}}{\mathrm{cos}{\mathrm{\theta }}_{1}\mathrm{cos}{\mathrm{\theta }}_{2}}\\ =\mathrm{tan}{\mathrm{\theta }}_{2}-\mathrm{tan}{\mathrm{\theta }}_{1}\end{array}$

Similarly, we can solve other terms which will be  Register to Get Free Mock Test and Study Material

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