Question

Problem D Find the exact value of each expression. If the answer is an angle, the angle must be in radians. 1) sec tan-- V3 2) cos tan 2 5 I 8 4) tan sin 3) sintan 17 6) cos- 5) cos(sin

Answer 1

1.
We know that,
$\tan^{-1} \left(\sqrt{3}\right)=\frac{\pi }{3}$
Then,
$\sec\left(\tan^{-1}\sqrt{3}\right) = \sec \frac{\pi}{3} = \frac{1}{\cos \frac{\pi}{3}} = \frac{1}{\frac{1}{2} }= 2$

2.
$\cos \tan ^{-1}\frac{2}{5}\:$

/1+ Use the identity: cos (tan-1 (z)) 1+2

$\cos \tan ^{-1}\frac{2}{5}\: = =\frac{\sqrt{1+\left(\frac{2}{5}\right)^2}}{1+\left(\frac{2}{5}\right)^2}= {\color{Blue} \frac{5\sqrt{29}}{29}} = {\color{Blue} 0.9284}$

3.
$\tan \left(-\:\frac{\pi }{4}\right) = -\tan \left(\:\frac{\pi }{4}\right) = -1$
$\sin ^{-1}\left(\tan \left(-\:\frac{\pi }{4}\right)\right) = \sin^{-1}(-1) = \sin^{-1}\sin \frac{-\pi}{2} = \frac{-\pi}{2} ={\color{Blue} \frac{-3.14}{2} = -1.57}$

4.
$\tan \sin ^{-1}\frac{8}{17}$
$\mathrm{Use\:the\:identity:}\:\tan \left(\sin^{-1} \left(x\right)\right)=\frac{x\sqrt{1-x^2}}{1-x^2}$
$\tan \sin ^{-1}\frac{8}{17} = =\frac{\left(\frac{8}{17}\right)\sqrt{1-\left(\frac{8}{17}\right)^2}}{1-\left(\frac{8}{17}\right)^2} ={\color{Blue} \frac{8}{15} =0.5333 }$

5.
$\cos ^{-1}\sin\left ( -\frac{\pi }{4} \right )$
$\mathrm{Use\:the\:\:identity}:\quad \sin \left(x\right)=\cos \left(\frac{\pi }{2}-x\right)$
$\\\cos ^{-1}\sin\left ( -\frac{\pi }{4} \right ) = \cos^{-1}\cos \left(\frac{\pi }{2}-\left(-\frac{\pi }{4}\right)\right) \\ = \cos^{-1}\cos\frac{3\pi }{4} ={\color{Blue} \frac{3\pi }{4} = \frac{3*3.14}{4} = \frac{9.42}{4} =2.355}$

6.
$\tan \frac{\pi}{4} = 1$
$\cos ^{-1}\tan \frac{\pi }{4} = \cos^{-1}1 = \cos^{-1}\cos 0 ={\color{Blue} 0}$


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