# trigonometry (3)

• October 7th 2009, 01:37 AM
thereddevils
trigonometry (3)
If $0 , and $0 , prove that

(1) $
\frac{1}{\cos^2 A}+\frac{1}{\sin^2 A\sin^2 B\cos^2 B}\geq 9
$

(2) Prove that $sin^2 A + sin^2 B > sin A + sin B + sin Asin B -1$
• October 7th 2009, 09:50 AM
red_dog
1) We have

$\frac{1}{\sin^2A\sin^2B\cos^2B}=\frac{\sin^2B+\cos ^2B}{\sin^2A\sin^2B\cos^2B}=\frac{1}{\sin^2A\cos^2 B}+\frac{1}{\sin^2A\sin^2B}$

Then we have to prove that

$\frac{1}{\cos^2A}+\frac{1}{\sin^2A\cos^2B}+\frac{1 }{\sin^2A\sin^2B}\geq 9$

Use the inequality $(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\r ight)\geq 9$

for $a=\cos^2A, \ b=sin^2A\cos^2B, \ c=\sin^2A\sin^2B$

using that

$a+b+c=\cos^2A+\sin^2A\cos^2B+\sin^2A\sin^2B=$

$=\cos^2A+\sin^2A(\sin^2B+\cos^2B)=\cos^2A+\sin^2A= 1$
• October 8th 2009, 04:38 AM
Quote:

Originally Posted by thereddevils
If $0 , and $0 , prove that

(1) $
\frac{1}{\cos^2 A}+\frac{1}{\sin^2 A \sin^2 B \cos^2 B} \geq 9
$

(2) Prove that $sin^2 A + sin^2 B > sin A + sin B + sin Asin B -1$

For (1) $\sec^2 A + \text{cosec}^2 A \, \text{cosec}^2 B \, \sec^2 B$

$=(1+\tan^2 A)+(1+\cot^2 A)(1+\cot^2 B)(1+tan^2 B)$

$=(1+\tan^2 A)+(1+\cot^2 A)(2+\tan^2 B+\cot^2 B)$

Consider $(\tan B-\cot B)^2\geq 0$

$\tan^2 B+\cot^2 B \geq 2$

Then $(\tan A-2\cot A)^2\geq 0$

$\tan^2 A+4\cot^2 A\geq 4$

continuing

$=(1+\tan^2 A)+(1+\cot^2 A)(4)$

$=5+\tan^2 A+4\cot^2 A$

$=9$

this inequality is at least 9 . Hence , proved .