If theta is an acute angle and tan(4theta - 5 | vedclass

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Question

(A) Given the equation: $	an(4	heta - 50^{\circ}) = \cot(50^{\circ} - 	heta)$.We know the trigonometric identity: $	an(A) = \cot(90^{\circ} - A)$.

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(A) Given the equation: $	an(4	heta - 50^{\circ}) = \cot(50^{\circ} - 	heta)$.We know the trigonometric identity: $	an(A) = \cot(90^{\circ} - A)$.Applying this identity to the left side,we get: $	an(4	heta - 50^{\circ}) = \cot(90^{\circ} - (4	heta - 50^{\circ}))$.Simplifying the expression inside the cotangent: $90^{\circ} - 4	heta + 50^{\circ} = 140^{\circ} - 4	heta$.So,$\cot(140^{\circ} - 4	heta) = \cot(50^{\circ} - 	heta)$.Equating the angles: $140^{\circ} - 4	heta = 50^{\circ} - 	heta$.Rearranging the terms: $140^{\circ} - 50^{\circ} = 4	heta - 	heta$.$90^{\circ} = 3	heta$.Therefore,$	heta = 30^{\circ}$.

If $\theta$ is an acute angle and $\tan(4\theta - 50^{\circ}) = \cot(50^{circ} - \theta)$,then the value of $\theta$ in degrees is:

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