Question

please fast solve my question.....✌️✌️​

Answer 1

GIVEN :–

• $\bf a_{1},a_{2},a_{3},......,a_{n}$ are in A.P. with Common difference 'd'.

TO FIND :–

• Value of

$\\ \implies \bf \tan \bigg[ \tan^{ - 1} \bigg( \frac{d}{1 +a_{1}a_{2}} \bigg) + \tan^{ - 1} \bigg( \frac{d}{1 +a_{2}a_{3}} \bigg) + \tan^{ - 1} \bigg( \frac{d}{1 +a_{3}a_{4}} \bigg) + ..... +\tan^{ - 1} \bigg( \frac{d}{1 +a_{n - 1}a_{n}} \bigg)\bigg]=?$

SOLUTION :–

• Let the function –

$\\ \implies \bf y =\tan \bigg[ \tan^{ - 1} \bigg( \frac{d}{1 +a_{1}a_{2}} \bigg) + \tan^{ - 1} \bigg( \frac{d}{1 +a_{2}a_{3}} \bigg) + \tan^{ - 1} \bigg( \frac{d}{1 +a_{3}a_{4}} \bigg) + ..... +\tan^{ - 1} \bigg( \frac{d}{1 +a_{n - 1}a_{n}} \bigg)\bigg]$

• If terms $\bf a_{1},a_{2},a_{3},......,a_{n}$ are in A.P. then we should write this as –

$\\ \implies\bf a_{2} = a_{1} + d$

$\\ \implies\bf a_{3} = a_{2} + d$

$\\ \implies\bf a_{n} = a_{n-1} + d$

• So that –

$\\ \implies\bf d = a_{2} - a_{1}$

$\\ \implies\bf d = a_{3} - a_{2}$

$\\ \implies\bf d = a_{n} - a_{n-1}$

• Put in equation –

$\\ \implies \bf y =\tan \bigg[ \tan^{ - 1} \bigg( \dfrac{a_{2} - a_{1}}{1 +a_{1}a_{2}} \bigg) + \tan^{ - 1} \bigg( \dfrac{a_{3} - a_{2}}{1 +a_{2}a_{3}} \bigg) + \tan^{ - 1} \bigg( \dfrac{a_{4} -a_{3}}{1 +a_{3}a_{4}} \bigg) + ..... +\tan^{ - 1} \bigg( \dfrac{a_{n} - a_{n-1}}{1 +a_{n - 1}a_{n}} \bigg)\bigg]$

• Using identity –

$\\ \implies \bf \tan^{ - 1} \bigg( \dfrac{a -b}{1 +ab} \bigg) = \tan^{ - 1} (a) - \tan^{ - 1} (b)$

$\\ \implies \bf y =\tan[\tan^{ - 1} ( a_{2} ) - \tan^{ - 1} (a_{1})+\tan^{ - 1} ( a_{3} ) - \tan^{ - 1} (a_{2})+\tan^{ - 1} ( a_{4} ) - \tan^{ - 1} (a_{3})+ ..... +\tan^{ - 1} ( a_{n} ) - \tan^{ - 1} (a_{n - 1})]$

$\\ \implies \bf y =\tan[\tan^{ - 1} ( a_{n} ) - \tan^{ - 1} (a_{1})]$

$\\ \implies \bf y =\tan \bigg[\tan^{ - 1} \bigg( \dfrac{a_{n} - a_{1}}{1 +a_{1}a_{n}} \bigg)\bigg]$

$\\ \large\implies{ \boxed{\bf y =\dfrac{a_{n} - a_{1}}{1 +a_{1}a_{n}}}}$

Answer 2

Answer:

GIVEN :–

• \bf a_{1},a_{2},a_{3},......,a_{n}a1,a2,a3,......,an are in A.P. with Common difference 'd'.

TO FIND :–

• Value of

\begin{gathered}\\ \implies \bf \tan \bigg[ \tan^{ - 1} \bigg( \frac{d}{1 +a_{1}a_{2}} \bigg) + \tan^{ - 1} \bigg( \frac{d}{1 +a_{2}a_{3}} \bigg) + \tan^{ - 1} \bigg( \frac{d}{1 +a_{3}a_{4}} \bigg) + ..... +\tan^{ - 1} \bigg( \frac{d}{1 +a_{n - 1}a_{n}} \bigg)\bigg]=?\\\end{gathered}⟹tan[tan−1(1+a1a2d)+tan−1(1+a2a3d)+tan−1(1+a3a4d)+.....+tan−1(1+an−1and)]=?

SOLUTION :–

• Let the function –

\begin{gathered}\\ \implies \bf y =\tan \bigg[ \tan^{ - 1} \bigg( \frac{d}{1 +a_{1}a_{2}} \bigg) + \tan^{ - 1} \bigg( \frac{d}{1 +a_{2}a_{3}} \bigg) + \tan^{ - 1} \bigg( \frac{d}{1 +a_{3}a_{4}} \bigg) + ..... +\tan^{ - 1} \bigg( \frac{d}{1 +a_{n - 1}a_{n}} \bigg)\bigg] \\\end{gathered}⟹y=tan[tan−1(1+a1a2d)+tan−1(1+a2a3d)+tan−1(1+a3a4d)+.....+tan−1(1+an−1and)]

• If terms \bf a_{1},a_{2},a_{3},......,a_{n}a1,a2,a3,......,an are in A.P. then we should write this as –

\begin{gathered}\\ \implies\bf a_{2} = a_{1} + d\\\end{gathered}⟹a2=a1+d

\begin{gathered}\\ \implies\bf a_{3} = a_{2} + d\\\end{gathered}⟹a3=a2+d

\begin{gathered}\\ \implies\bf a_{n} = a_{n-1} + d\\\end{gathered}⟹an=an−1+d

• So that –

\begin{gathered}\\ \implies\bf d = a_{2} - a_{1} \\\end{gathered}⟹d=a2−a1

\begin{gathered}\\ \implies\bf d = a_{3} - a_{2}\\\end{gathered}⟹d=a3−a2

\begin{gathered}\\ \implies\bf d = a_{n} - a_{n-1} \\\end{gathered}⟹d=an−an−1

• Put in equation –

\begin{gathered}\\ \implies \bf y =\tan \bigg[ \tan^{ - 1} \bigg( \dfrac{a_{2} - a_{1}}{1 +a_{1}a_{2}} \bigg) + \tan^{ - 1} \bigg( \dfrac{a_{3} - a_{2}}{1 +a_{2}a_{3}} \bigg) + \tan^{ - 1} \bigg( \dfrac{a_{4} -a_{3}}{1 +a_{3}a_{4}} \bigg) + ..... +\tan^{ - 1} \bigg( \dfrac{a_{n} - a_{n-1}}{1 +a_{n - 1}a_{n}} \bigg)\bigg] \\\end{gathered}⟹y=tan[tan−1(1+a1a2a2−a1)+tan−1(1+a2a3a3−a2)+tan−1(1+a3