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

$\sin (45^{\circ}+\theta)-\cos (45^{\circ}-\theta)$ is equal to

$\cot \theta \cdot \tan \theta = \dots$

Given that $\alpha + \beta = 90^{\circ}$,show that $\sqrt{\cos \alpha \operatorname{cosec} \beta - \cos \alpha \sin \beta} = \sin \alpha$.

If $\sin 5 \theta = \cos 5 \theta$,then the value of $\theta$ is ..........

Write 'True' or 'False' and justify your answer.
The value of $2 \sin \theta$ can be $(a + \frac{1}{a}),$ where $a$ is a positive number,and $a \neq 1$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module