The three sides of the triangle are: sin a, cos a and sqrt(1+sinacosa) (Given, a = alpha)
Let A = sin a, B = cos a and C = sqrt(1+sinacosa)
It is clear, that C > A and C > B, therefore the greatest side of the triangle is C and we also know that the angle opposite to the greatest side is the greatest angle of the triangle. Therefore, the greatest angle in this case is ‘c’.
Now, using the trigonometric identity, cos C = (A^2 + B^2 – C^2)/2AB, we have:
cos c = {(sin a)^2 + (cos a)^2 – (sqrt(1+sinacosa))^2} / (2sinacos a)
cos c = {(sin a)^2 + (cos a)^2 – (1+sinacosa)} / (2sin acos a)
cos c = (1 – 1 – sinacosa) / (2sinacos a) [Because we know that, (sin a)^2 + (cos a)^2 =1]
cos c = - sinacosa / 2sinacosa
cos c = -1/2
c = 120 degree (Answer)
Therefore, it’s an obtuse angle triangle and the greatest angle is 120 degree.
Question : Simplify the following. $\frac{\sin^3 \alpha+\cos^3 \alpha}{\sin \alpha+\cos \alpha}$
Option 1: $1+\sin \alpha \cos \alpha$
Option 2: $\tan \alpha$
Option 3: $1-\sin \alpha \cos \alpha$
Option 4: $\sec \alpha$
Question : What is $\sin \alpha - \sin\beta$?
Option 1: $2 \cos \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2}$
Option 2: $2 \sin \frac{\alpha+\beta}{2} \sin \frac{\alpha-\beta}{2}$
Option 3: $2 \cos \frac{\alpha-\beta}{2} \sin \frac{\alpha+\beta}{2}$
Option 4: $2 \cos \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2}$
Question : If $\cos^{2}\alpha-\sin^{2}\alpha=\tan^{2}\beta$, then the value of $\cos^{2}\beta-\sin^{2}\beta$ is:
Option 1: $\cot^{2}\alpha$
Option 2: $\cot^{2}\beta$
Option 3: $\tan^{2}\alpha$
Option 4: $\tan^{2}\beta$
Question : If A is an acute angle, the simplified form of $\frac{\cos (\pi-A) \cdot \cot \left(\frac{\pi}{2}+A\right) \cos (-A)}{\tan (\pi+A) \tan \left(\frac{3 \pi}{2}+A\right) \sin (2 \pi-A)}$ is:
Option 1: $ \cos^2 A$
Option 2: $\sin A$
Option 3: $\sin^2 A$
Option 4: $\cos A$
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