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Arithmetic Progressions — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsArithmetic Progressions5 Marks
Question
If $\theta_1,\ \theta_2,\ \theta_3,\ ...\theta_\text{n}$ are in AP. whose common difference is d, show thet $\sec\theta_1\sec\theta_2+\sec\theta_2\sec\theta_3+...+\sec\theta_{\text{n}-1}\sec\theta_\text{n}=\frac{\theta_\text{n}-\tan\theta_1}{\sin\text{d}}$
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Answer

$\sec\theta_1\sec\theta_2+\sec\theta_2\sec\theta_3+...+\sec\theta_{\text{n}-1}\sec\theta_\text{n}=\frac{\theta_\text{n}-\tan\theta_1}{\sin\text{d}}$
$\theta_2-​​\theta_1=\theta_3-\theta_2=......=\text{d}$
$\sec\theta-1\sec\theta_2=\frac{1}{\cos\theta_1\cos\theta_2}=\frac{\sin\text{d}}{\sin\text{d}(\cos\theta_1\cos\theta_2)}$
$=\frac{\sin(\theta_2-\theta_1)}{\sin\text{d}(\cos\theta_1\cos\theta_2)}$
$=\frac{\sin\theta_2\cos\theta_1-\cos\theta_2\sin\theta_1}{\sin\text{d}(\cos\theta_1\cos\theta_2)}$
$=\frac{1}{\sin\text{d}}\Big[\frac{\sin\theta_2\cos\theta_1}{(\cos\theta_1\cos\theta_2)}-\frac{\cos\theta_2\sin\theta_1}{(\cos\theta_1\cos\theta_2)}\Big]$
$=\frac{1}{\sin\text{d}}[\tan\theta_2-\tan\theta_1]$
Similaely, $\sec\theta_2\sec\theta_3=\frac{1}{\sin\text{d}}[\tan\theta_3\tan\theta-2]$
If we add up all terms, we get
$=\frac{1}{\sin\text{d}}[\tan\theta_2-\tan\theta_1+\tan\theta_3-\tan\theta-2+......+\tan\theta_\text{n}=\tan\theta_{\text{n}-1}]$
$=\frac{1}{\sin\text{d}}[\tan\theta_\text{n}-\tan\theta_1]$
Hence proved.

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