Tan 23,,tan 42,, tan 48,, tan 67=ldots ldots ldots ldots .

English

Gujarati

Hindi

8. Introduction to Trigonometry

hard

$\tan 23\,\,\tan 42\,\, \tan 48\,\, \tan 67=\ldots \ldots \ldots \ldots .$

A

$4$

B

$3$

C

$2$

D

$1$

Solution

$\tan 48=\cot (90-48)=\cot 42$ and $\tan 67=\cot (90-67)=\cot 23$

Now, tan $23 \tan \,\,42 \tan \,\,48 \tan 67=\tan 23 \,\,\tan 42\, \,\cot 42\, \,\cot 23$

$=(\tan 23\, \,\cot 23)(\tan 42\,\, \cot 42)$

$=1 \times 1=1$

Standard 10

Mathematics

Share

Similar Questions

$\tan \theta=\frac{5}{12},$ then $\cos \theta=\ldots \ldots \ldots \ldots$

easy

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

medium

$\cos \theta=\frac{15}{17},$ then the value of $\operatorname{cosec} \theta+\cot \theta $ is ………

easy

$(\sin 80+\cos 10)(\sin 80-\cos 10)=\ldots \ldots \ldots$

easy

$\sin ^{2} 35+\cos ^{2} \theta=1,$ then $\theta=\ldots \ldots \ldots \ldots$

easy