If -1- -2- -3- -n are in A-P- whose common difference is d- show that sec -1-2-2-3-n-1 sec -n-tan-n-tan-1-sin d

As θ_1, θ_2, θ_3, ……θ_n are in A.P. 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 =1/cos θ_ ...

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Byju's Answer

Standard XII

Mathematics

Trigonometric Ratios of Multiples of an Angle

If θ1, θ2, θ3...

Question

If $θ_{1}, θ_{2}, θ_{3}, \dots \dots θ_{n}$ are in A.P, whose common difference is d, show that $s e c θ_{1} s e c θ_{2} + s e c θ_{2} s e c θ_{3} + \dots \dots + s e c θ_{n - 1} s e c θ_{n} = \frac{t a n θ_{n} - t a n θ_{1}}{s i n d}$

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Solution

As $θ_{1}, θ_{2}, θ_{3}, \dots \dots θ_{n}$ are in A.P.
So, $d = θ_{2} - θ_{1} = θ_{3} - θ_{2} = \dots \dots = θ_{n} - θ_{n - 1} \dots \dots (i)$
Now,
LHS $s e c θ_{1} s e c θ_{2} + s e c θ_{2} s e c θ_{3} + \dots \dots + s e c θ_{n - 1} s e c θ_{n}$
$= \frac{1}{c o s θ_{1} c o s θ_{2}} + \frac{1}{c o s θ_{2} c o s θ_{3}} + \dots \dots + \frac{1}{c o s θ_{n - 1} c o s θ_{n}} = \frac{1}{s i n d} [\frac{s i n d}{c o s θ_{1} c o s θ_{2}} + \frac{s i n d}{c o s θ_{2} c o s θ_{3}} + \dots \dots + \frac{s i n d}{c o s θ_{n - 1} c o s θ_{n}}]$
$= \frac{1}{s i n d} [\frac{s i n (θ_{2} - θ_{1})}{c o s θ_{1} c o s θ_{2}} + \frac{s i n (θ_{3} - θ_{2})}{c o s θ_{2} c o s θ_{3}} + \dots \dots + \frac{s i n (θ_{n} - θ_{n - 1})}{c o s θ_{n - 1} c o s θ_{n}}]$ [Using (i)]
$= \frac{1}{s i n d} [\frac{s i n θ_{2} c o s θ_{1} - c o s θ_{2} s i n θ_{1}}{c o s θ_{1} c o s θ_{2}} + \frac{s i n θ_{3} c o s θ_{2} - c o s θ_{3} s i n θ_{2}}{c o s θ_{2} c o s θ_{3}} + \dots \dots + \frac{s i n θ_{n} c o s θ_{n - 1} - c o s θ_{n} s i n θ_{n - 1}}{c o s θ_{n - 1} c o s θ_{n}}]$
$= \frac{1}{s i n d} [\frac{s i n θ_{2} c o s θ_{1}}{c o s θ_{1} c o s θ_{2}} - \frac{c o s θ_{2} s i n θ_{1}}{c o s θ_{1} c o s θ_{2}} + \frac{s i n θ_{3} c o s θ_{2}}{c o s θ_{2} c o s θ_{3}} - \frac{c o s θ_{3} s i n θ_{2}}{c o s θ_{2} c o s θ_{3}} + \dots \dots + \frac{s i n θ_{n} c o s θ_{n - 1}}{c o s θ_{n - 1} c o s θ_{n}} - \frac{c o s θ_{n} s i n θ_{n - 1}}{c o s θ_{n - 1} c o s θ_{n}}]$
$= \frac{1}{s i n d} [t a n θ_{2} - t a n θ_{1} + t a n θ_{3} - t a n θ_{2} + \dots \dots + t a n θ_{n} - t a n θ_{n - 1}]$
$= \frac{1}{s i n d} [- t a n θ_{1} + t a n θ_{n}] = \frac{t a n θ_{n} - t a n θ_{1}}{s i n d}$
= RHS

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