For acute angle theta, if cos theta = sin the | vedclass

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(D) Given $\cos 	heta = \sin 	heta.$Dividing both sides by $\cos 	heta,$ we get $\frac{\cos 	heta}{\cos 	heta} = \frac{\sin 	heta}{\cos 	heta}.

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$\frac{7}{2}$

Vedclass · Answer

(D) Given $\cos 	heta = \sin 	heta.$Dividing both sides by $\cos 	heta,$ we get $\frac{\cos 	heta}{\cos 	heta} = \frac{\sin 	heta}{\cos 	heta}.$Therefore,$1 = 	an 	heta.$Since $	heta$ is an acute angle,$	heta = 45^{\circ}.$Now,substitute $	heta = 45^{\circ}$ into the expression $2 	an^{2} 	heta + \sin^{2} 	heta + 1.$$2 	an^{2} 45^{\circ} + \sin^{2} 45^{\circ} + 1 = 2(1)^{2} + (\frac{1}{\sqrt{2}})^{2} + 1.$$= 2(1) + \frac{1}{2} + 1.$$= 3 + \frac{1}{2} = \frac{7}{2}.$

$1. \cos \theta$	$a. \frac{\cos \theta}{\sin \theta}$
$2. \tan \theta$	$b. \frac{1}{\csc \theta}$
$3. \cot \theta$	$c. \frac{1}{\sec \theta}$
$4. \sin \theta$	$d. \frac{1}{\cot \theta}$
	$e. \sin \theta \cdot \cos \theta$

For acute angle $\theta,$ if $\cos \theta = \sin \theta,$ then $2 \tan^{2} \theta + \sin^{2} \theta + 1 = \ldots$

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