tan 23^{circ} tan 42^{circ} tan 48^{circ} tan | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) We know that $	an 	heta = \cot(90^{\circ} - 	heta)$.Using this identity:$	an 48^{\circ} = 	an(90^{\circ} - 42^{\circ}) = \cot 42^{\circ}$$	a

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(D) We know that $	an 	heta = \cot(90^{\circ} - 	heta)$.Using this identity:$	an 48^{\circ} = 	an(90^{\circ} - 42^{\circ}) = \cot 42^{\circ}$$	an 67^{\circ} = 	an(90^{\circ} - 23^{\circ}) = \cot 23^{\circ}$Now,substitute these values into the expression:$	an 23^{\circ} 	an 42^{\circ} 	an 48^{\circ} 	an 67^{\circ} = 	an 23^{\circ} 	an 42^{\circ} \cot 42^{\circ} \cot 23^{\circ}$Rearranging the terms:$= (	an 23^{\circ} \cot 23^{\circ}) 	imes (	an 42^{\circ} \cot 42^{\circ})$Since $	an 	heta 	imes \cot 	heta = 1$:$= 1 	imes 1 = 1$

$\tan 23^{\circ} \tan 42^{\circ} \tan 48^{\circ} \tan 67^{\circ} = \ldots \ldots \ldots \ldots .$

Similar Questions

If $\tan A = \cot B$,then $A + B = \ldots$

If $\tan \theta + \sec \theta = l$,then prove that $\sec \theta = \frac{l^{2} + 1}{2l}$.

State whether the following is 'True' or 'False' and justify your answer:
If $\cos A + \cos^2 A = 1$,then $\sin^2 A + \sin^4 A = 1$.

Given that $\sin \theta = \frac{a}{b},$ then $\cos \theta$ is equal to

Prove that $(\sin^{4} \theta - \cos^{4} \theta + 1) \operatorname{cosec}^{2} \theta = 2$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module