If _1, _2, _3,.... _n are in A.P., whose common difference is d, show that _1 _2+ _2+ _3+ + _ n-1 _n= _n- _1

Since, _1, _2, _3, _n are in A.P. _2- _1= _3- _2= _n- _ n-1 =d Now we have to prove that _1 _2+ _2 _3+ + _ n-1 _n = _n- _1 L.H.S. [ _1 _2+ _2 _3+ + _ n-1 _n ] Taking only [ _1 _2] = [1 1 _2 ] = ( _2- _1) 1 _1 _2 =1 [ _2 _1- _2 _1 _1 _2 ] =1 [ _2 _1 _1 _2 - _2 _1 _1 _2 ] =1 [ _2- _1] Similarly we can solve other terms which will be =1 [ _3- _2] and 1 [ _4- _3] Here L.H.S. =1 [ _2- _1+ _3- _2+ + _n- _ n-1 ] =1 [- _1+ _n]= _n- _1 R.H.S.

If _1, _2, _3,.... _n are in A.P., whose common difference is d, …

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Question

If $\theta_1,\theta_2,\theta_3,....\theta_\text{n}$ are in A.P., whose common difference is d, show that $\sec\theta_1\cdot\sec\theta_2+\sec\theta_2+\sec\theta_3+\dots+\sec\theta_{\text{n}-1}\cdot\sec\theta_\text{n}=\frac{\tan\theta_\text{n}-\tan\theta_1}{\sin\text{d}}$

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Answer

Since, $\theta_1,\theta_2,\theta_3,\cdots\theta_\text{n}$ are in A.P.
$\therefore\ \theta_2-\theta_1=\theta_3-\theta_2=\theta_\text{n}-\theta _{\text{n}-1}=\text{d}$
Now we have to prove that
$\sec\theta_1\cdot\sec\theta_2+\sec\theta_2\cdot\sec\theta_3+\cdots+\sec\theta_{\text{n}-1}\cdot\sec\theta_\text{n}$
$=\frac{\tan\theta_\text{n}-\tan\theta_1}{\sin\text{d}}\text{L.H.S}.$
$\Rightarrow\frac{\sin\text{d}}{\sin\text{d}}\big[\sec\theta_1\cdot\sec\theta_2+\sec\theta_2\sec\theta_3+\cdots+\sec\theta_{\text{n}-1}\cdot\sec\theta_\text{n}\big]$
Taking only $ \frac{\sin\text{d}[\sec\theta_1\cdot\sec\theta_2]}{\sin\text{d}}=\frac{\sin\text{d}\Big[\frac{1}{\cos\theta}\cdot\frac{1}{\cos\theta _2}\Big]}{\sin\text{d}}$
$=\frac{\sin(\theta_2-\theta_1)}{\sin\text{d}}\cdot\frac{1}{\cos\theta_1\cos\theta_2}$
$=\frac{1}{\sin\text{d}}\Big[\frac{\sin\theta_2\cos\theta_1-\cos\theta_2\sin\theta_1}{\cos\theta_1\cos\theta_2}\Big]$
$=\frac{1}{\sin\text{d}}\Big[\frac{\sin\theta_2\cos\theta_1}{\cos\theta_1\cos\theta_2}-\frac{\cos\theta_2\sin\theta_1}{\cos\theta_1\cos\theta_2}\Big]$
$=\frac{1}{\sin\text{d}}[\tan\theta_2-\tan\theta_1]$
Similarly we can solve other terms which will be
$=\frac{1}{\sin\text{d}}[\tan\theta_3-\tan\theta_2]$ and $\frac{1}{\sin\text{d}}[\tan\theta_4-\tan\theta_3]$
Here L.H.S. $=\frac{1}{\sin\text{d}}\Big[\tan\theta_2-\tan\theta_1+\tan\theta_3-\tan\theta_2+\cdots+\tan \theta_\text{n}-\tan\theta_{\text{n}-1}\Big]$
$=\frac{1}{\sin\text{d}}[-\tan\theta_1+\tan\theta_\text{n}]=\frac{\tan\theta_\text{n}-\tan\theta_1}{\sin\text{d}}\text{R.H.S.}$