If - 1- - 2- - 3- - - n are in AP- whose common difference is d- then show thatsec - 1 - 2- 2 - 3- n-1 - n -tan- n -tan - 1 sind NCERT EXEMPLAR

As, #160; #952;1, #160; #952;2, #160; #952;3, #160;..., #160; #952;n #160;are #160;in #160;APSo, #160;d= #952;2 #952;1= #952;3 ...

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Standard XII

Mathematics

Improper Integrals

If θ 1, θ 2, ...

Question

$If θ_{1}, θ_{2}, θ_{3}, . . ., θ_{n} are in AP, whose common difference is d, then show that \sec θ_{1} \sec θ_{2} + \sec θ_{2} \sec θ_{3} + . . . + \sec θ_{n - 1} \sec θ_{n} = \frac{\tan θ_{n} - \tan θ_{1}}{\sin d} [NCERT EXEMPLAR]$

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Solution

$As, θ_{1}, θ_{2}, θ_{3}, . . ., θ_{n} are in AP So, d = θ_{2} - θ_{1} = θ_{3} - θ_{2} = . . . = θ_{n} - θ_{n - 1} . . . . . (i) Now, LHS = \sec θ_{1} \sec θ_{2} + \sec θ_{2} \sec θ_{3} + . . . + \sec θ_{n - 1} \sec θ_{n} = \frac{1}{\cos θ_{1} \cos θ_{2}} + \frac{1}{\cos θ_{2} \cos θ_{3}} + . . . + \frac{1}{\cos θ_{n - 1} \cos θ_{n}} = \frac{1}{\sin d} [\frac{\sin d}{\cos θ_{1} \cos θ_{2}} + \frac{\sin d}{\cos θ_{2} \cos θ_{3}} + . . . + \frac{\sin d}{\cos θ_{n - 1} \cos θ_{n}}] = \frac{1}{\sin d} [\frac{\sin (θ_{2} - θ_{1})}{\cos θ_{1} \cos θ_{2}} + \frac{\sin (θ_{3} - θ_{2})}{\cos θ_{2} \cos θ_{3}} + . . . + \frac{\sin (θ_{n} - θ_{n - 1})}{\cos θ_{n - 1} \cos θ_{n}}] [Using (i)] = \frac{1}{\sin d} [\frac{\sin θ_{2} \cos θ_{1} - \cos θ_{2} \sin θ_{1}}{c o s θ_{1} c o s θ_{2}} + \frac{\sin θ_{3} \cos θ_{2} - \cos θ_{3} \sin θ_{2}}{c o s θ_{2} c o s θ_{3}} + . . . + \frac{\sin θ_{n} \cos θ_{n - 1} - \cos θ_{n} \sin θ_{n - 1}}{c o s θ_{n - 1} c o s θ_{n}}] = \frac{1}{\sin d} [\frac{\sin θ_{2} \cos θ_{1}}{\cos θ_{1} \cos θ_{2}} - \frac{\cos θ_{2} \sin θ_{1}}{\cos θ_{1} \cos θ_{2}} + \frac{\sin θ_{3} \cos θ_{2}}{\cos θ_{2} \cos θ_{3}} - \frac{\cos θ_{3} \sin θ_{2}}{\cos θ_{2} \cos θ_{3}} + . . . + \frac{\sin θ_{n} \cos θ_{n - 1}}{\cos θ_{n - 1} \cos θ_{n}} - \frac{\cos θ_{n} \sin θ_{n - 1}}{\cos θ_{n - 1} \cos θ_{n}}] = \frac{1}{\sin d} [\tan θ_{2} - \tan θ_{1} + \tan θ_{3} - \tan θ_{2} + . . . + \tan θ_{n} - \tan θ_{n - 1}] = \frac{1}{\sin d} [- \tan θ_{1} + \tan θ_{n}] = \frac{\tan θ_{n} - \tan θ_{1}}{\sin d} = RHS$

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If θ1, θ2, θ3, ..., θn are in AP, whose common difference is d, then show thatsecθ1secθ2+secθ2secθ3+...+secθn-1secθn=tanθn-tanθ1sind NCERT EXEMPLAR

$If θ_{1}, θ_{2}, θ_{3}, . . ., θ_{n} are in AP, whose common difference is d, then show that \sec θ_{1} \sec θ_{2} + \sec θ_{2} \sec θ_{3} + . . . + \sec θ_{n - 1} \sec θ_{n} = \frac{\tan θ_{n} - \tan θ_{1}}{\sin d} [NCERT EXEMPLAR]$