Trigonometry> if sec (a+b),sec a, sec (a-b) are in a.P....

if sec (a+b),sec a, sec (a-b) are in a.P. then the value of cosa secb/2

laxman , 11 Years ago

Grade 11

1 Answers

Jitender Singh

Last Activity: 11 Years ago

Ans:
Hello Student,
Please find answer to your question below

sec(a+b), seca, sec(a-b) are in AP, we have
$seca - sec(a+b) = sec(a-b) - seca$
$seca + seca = sec(a-b) + sec(a+b)$
$2seca = \frac{1}{cos(a-b)} + \frac{1}{cos(a+b)}$
$2seca = \frac{cos(a+b)+cos(a-b)}{cos(a-b).cos(a+b)}$
$2seca = \frac{cosa.cosb-sina.sinb+cosa.cosb+sina.sinb}{(cosa.cosb+sina.sinb).(cosa.cosb-sina.sinb)}$
$2seca = \frac{2cosa.cosb}{(cos^{2}a.cos^{2}b-sin^{2}a.sin^{2}b)}$
$seca = \frac{cosa.cosb}{(cos^{2}a.cos^{2}b-(1-cos^{2}a)(1-cos^{2}b)}$
$seca = \frac{cosa.cosb}{(cos^{2}a.cos^{2}b-1+cos^{2}a+cos^{2}b-cos^{2}a.cos^{2}b)}$
$seca = \frac{cosa.cosb}{(cos^{2}a+cos^{2}b-1)}$
$\frac{1}{cosa} = \frac{cosa.cosb}{(cos^{2}a+cos^{2}b-1)}$
$cos^{2}a+cos^{2}b-1 = cos^{2}a.cosb$
$cos^{2}a(1-cosb) =1-cos^{2}b$
$cos^{2}a(1-cosb) =(1-cosb)(1+cosb)$
$cos^{2}a =1+cosb$
$cos^{2}a = 2cos^{2}\frac{b}{2}$
$cos^{2}a.sec^{2}\frac{b}{2} = 2$
$cosa.sec\frac{b}{2} = \sqrt{2}$