8. Introduction to TrigonometryMedium

સાબિત કરો :

$(i)$ $\tan 48^{\circ} \tan 23^{\circ} \tan 42^{\circ} \tan 67^{\circ}=1$

$(ii)$ $\cos 38^{\circ} \cos 52^{\circ}-\sin 38^{\circ} \sin 52^{\circ}=0$

Solution

(i) $\tan 48^{\circ} \tan 23^{\circ} \tan 42^{\circ} \tan 67^{\circ}$

$=\tan \left(90^{\circ}-42^{\circ}\right) \tan \left(90^{\circ}-67^{\circ}\right) \tan 42^{\circ} \tan 67^{\circ}$

$=\cot 42^{*} \cot 67^{*} \tan 42^{*} \tan 67^{\circ}$

$=\left(\cot 42^{\circ} \tan 42^{\circ}\right)\left(\cot 67^{*} \tan 67^{\circ}\right)$

$=(1)(1)$

$=1$

(ii) $\cos 38^{\circ} \cos 52^{\circ}-\sin 38^{\circ} \sin 52^{\circ}$

$=\cos \left(90^{\circ}-52^{\circ}\right) \cos \left(90^{\circ}-38^{\circ}\right)-\sin 38^{\circ} \sin 52^{\circ}$

$=\sin 52^{*} \sin 38^{\circ}-\sin 38^{\circ} \sin 52^{*}$

$=0$

$\triangle$ $PQR$માં, $Q$ કાટખૂણો છે (જુઓ આકૃતિ). $PQ = 3$ સેમી અને $PR = 6$ સેમી હોય, તો $\angle QPR$ અને $\angle PRQ$ શોધો.

Difficult

Medium

Medium

Medium

Easy