Question

solve

Answer 1

Since tanA , tan B and tan C are in A.p
so tanA=tanB+tanC

Answer 2

sorry
2 tanB=tanA+tanC

Answer 3

use the identitites to solve it

Answer 4

how

Answer 5

this is abrute way, but there might be a more practical way to go about it
\[\frac{sina}{cosa}=\frac{sina}{cosa}+d(0)\]
\[\frac{sinb}{cosb}=\frac{sina}{cosa}+d(1)\]
\[\frac{sinc}{cosc}=\frac{sina}{cosa}+d(2)\]------------------------

\[\frac{sina}{cosa}\frac{cos^2a}{sina}=\frac{sina}{cosa}\frac{cos^2a}{sina}+d(0)\frac{cos^2a}{sina}\]
\[\frac{sinb}{cosb}\frac{cos^2a}{sina}=\frac{sina}{cosa}\frac{cos^2a}{sina}+d(1)\frac{cos^2a}{sina}\]
\[\frac{sinc}{cosc}\frac{cos^2a}{sina}=\frac{sina}{cosa}\frac{cos^2a}{sina}+d(2)\frac{cos^2a}{sina}\]-----------------------

\[cosa=cosa+d(0)\frac{cos^2a}{sina}\]
\[\frac{sinb}{cosb}\frac{cos^2a}{sina}=cosa+d(1)\frac{cos^2a}{sina}\]
\[\frac{sinc}{cosc}\frac{cos^2a}{sina}=cosa+d(2)\frac{cos^2a}{sina}\]

if you can show that those leftsides simplify to cosb and cosc ...

Answer 6

how ur first step came

Answer 7

thats just the definition of an AP

Answer 8

srry but i didnt get it

Answer 9

i havent gotten it either, that was just an attempt :) but i started out using the definijtion of an AP, each term is a equal to the one before it, plus some common addon.

a0 = n + d(0)
a1 = n + d(1)
a2 = n + d(2)
a3 = n + d(3)
...
ak = n + d(k)

Answer 10

scaling the sequence shouldnt change the AP nature of it

pa0 = pn + pd(0)
pa1 = pn + pd(1)
pa2 = pn + pd(2)
pa3 = pn + pd(3)

Answer 11

A= {0,3,6,9,12,15,...}

3A= {0,9,18,27,36,45,...}

Answer 12

but the trick is in seeing of the tans can simplify to coss

Answer 13

plse can u solve this sum plsee

Answer 14

i cant, no. there are some things that im just not capable of doing at times :)

Answer 15

okay