Question

If cot A + cosec A = 1.5 , then show that cos A= 5/13..

Answer 1

reply ASAP please

Answer 2

\[1+\cot ^{2}A=\csc ^{2}A\]\[\csc ^{2}A-\cot ^{2}A=1\]\[\left( \csc A+\cot A \right)\left( \csc A-\cot A \right)=1\]\[\frac{3}{2}\left( \csc A-\cot A \right)=1\]\[\csc A-\cot A=\frac{2}{3}\]\[\left( \csc A-\cot A \right)+\left( \csc A+\cot A \right)=\frac{2}{3}+\frac{3}{2}\]\[\csc A=\frac{13}{12}\]\[\sin A=\frac{12}{13}\]\[\cos A=\sqrt{1-\sin ^{2}A}=\frac{5}{13}\]

Answer 3

thanks

Answer 4

excellent way ... i never had seen one that way.

Answer 5

I learned that way from my previous math competition problems :)

Answer 6

1+cot2A=csc2A
csc2A−cot2A=1
(cscA+cotA)(cscA−cotA)=1
32(cscA−cotA)=1
cscA−cotA=23
(cscA−cotA)+(cscA+cotA)=23+32
cscA=1312
sinA=1213
cosA=1−sin2A−−−−−−−−√=513

Answer 7

I like @exraven's approach. +1

Answer 8

However I would have done it by solving the quadratic.

Answer 9

\[ 13x^2 +8x-5=0 \text{ where } x= \cos x \]

Answer 10

same here ... the classic way.

Answer 11

Unless it's saves time (drastically) classic rules for me! :)