# trig identities help needed solving????

• May 9th 2012, 11:20 PM
koolaid123
trig identities help needed solving????
1/sec + tan = sec - tan
prove one side equals to another BY SOLVING ONE SIDE ONLY
• May 10th 2012, 12:00 AM
princeps
Re: trig identities help needed solving????
$\frac{1}{\sec \alpha+\tan \alpha}=\frac{1}{\frac{1}{\cos \alpha}+\frac{\sin \alpha}{\cos \alpha}}=\frac{\cos \alpha}{1+\sin \alpha} \cdot \frac{1-\sin \alpha}{1-\sin \alpha}=\frac{\cos \alpha \cdot(1-\sin \alpha)}{\cos^2 \alpha}=\frac{1-\sin \alpha}{\cos \alpha}=\frac{1}{\cos \alpha}-\tan \alpha=\sec \alpha-\tan \alpha$
• May 10th 2012, 01:40 AM
BobP
Re: trig identities help needed solving????
The LHS should be $1/(\sec\theta+\tan\theta).$ Multiply top and bottom by $(\sec\theta-\tan\theta)$.