Tan 48-tan 23-tan 42-tan 67-

The correct option is C 1⇒ tan 48^∘ tan 23^∘ tan 42^∘ tan 67^∘ =(tan 48^∘ tan 42^∘) (tan 23^∘ tan 67^∘) ={tan 48^∘. tan(90^∘ 48^∘)}{tan 23^∘. tan(90^∘ 23^∘)} ...

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Byju's Answer

Standard IX

Mathematics

Trigonometric Ratios of Complementary Angles

tan 48∘tan 23...

Question

$t a n 48^{\circ} t a n 23^{\circ} t a n 42^{\circ} t a n 67^{\circ} =$

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Solution

The correct option is C 1
$\Rightarrow t a n 48^{\circ} t a n 23^{\circ} t a n 42^{\circ} t a n 67^{\circ}$

$= (t a n 48^{\circ} t a n 42^{\circ}) (t a n 23^{\circ} t a n 67^{\circ})$

$= {t a n 48^{\circ} . t a n (90^{\circ} - 48^{\circ})} {t a n 23^{\circ} . t a n (90^{\circ} - 23^{\circ})}$

$= (t a n 48^{\circ} c o t 48^{\circ}) (t a n 23^{\circ} c o t 23^{\circ})$

$[∵ t a n (90^{\circ} - θ) = c o t θ]$

$= 1 \times 1 = 1$

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Similar questions

tan 48^{\circ} . tan 23^{\circ} . tan 42^{\circ} . tan 67^{\circ}

= ___

Show that $tan 48^{\circ} tan 23^{\circ} tan 42^{\circ} tan 67^{\circ} = 1$

Q. Show that :

t a n 48^{\circ} t a n 23^{\circ} t a n 42^{\circ} t a n 67^{\circ} = 1

The value of
$t a n 48^{\circ} . t a n 23^{\circ} . t a n 42^{\circ} . t a n 67^{\circ}$ =___

Q. Question 2 (i)
Show that :

t a n 48^{\circ} t a n 23^{\circ} t a n 42^{\circ} t a n 67^{\circ} = 1

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tan 48∘ tan 23∘ tan 42∘ tan 67∘=

The correct option is C 1⇒tan 48∘ tan 23∘ tan 42∘ tan 67∘ =(tan 48∘ tan 42∘)(tan 23∘ tan 67∘) ={tan 48∘.tan(90∘−48∘)}{tan 23∘.tan(90∘−23∘)} =(tan 48∘ cot 48∘)(tan 23∘ cot 23∘) [∵tan(90∘−θ)=cot θ] =1×1=1

$t a n 48^{\circ} t a n 23^{\circ} t a n 42^{\circ} t a n 67^{\circ} =$