Question

दिखाइए कि
(i) $tan 48^{o} tan 23^{o} tan 42^{o} tan 67^{o} = 1$
(ii) $cos 38^{o} cos 52^{o} - sin 38^{o} sin 52^{o} = 0$

Answer 1

Answer (i)

tan 48° tan 23° tan 42° tan 67° = 1

Take LHS

$\implies$tan 48° tan 42° tan 67° tan 23°

$\implies$tan 48° tan (90 - 48)° tan 67° tan (90-67)°

$\implies$tan 48° cot 48° tan 67° cot 67°

$\implies tan \: 48^{\circ} \: \dfrac{1}{tan\: 48^{\circ}} \: tan \: 67^{\circ} \: \dfrac{1}{tan \: 67^{\circ}}$

$\implies \bold{1}$

RHS

LHS = RHS (Hence proved)

Answer (ii)

$cos 38^{\circ} \: cos 52^{\circ} - sin 38^{\circ} \: sin 52^{o} \: = \: 0$

Take LHS

$\implies$ cos 38° cos (90 - 38)° - sin 38° sin (90 - 38)°

$\implies$ cos 38° sin 38° - sin 38° cos 38°

$\implies \bold{0}$

RHS

LHS = RHS ( Hence Proved)

$\bold{Thanks}$

Answer 2

(1) Tan48° × Tan23° × Tan42° × Tan67° = 1




=> LHS = Tan48° × Tan23° × Tan42° × Tan67°




=> Tan48° × Tan42° × Tan23° × Tan67°



=> Tan48 × Tan ( 90 - 48 ) × Tan23° × Tan(90-23 ).


=> Tan48 × Cot48 × Tan23° × Cot23° .



=> 1/ Cot48° × Cot48 × Tan23° × 1/ Tan23° .




=> 1



RHS = 1.




Hence,

LHS = RHS



(2) Cos38° × Cos52° - Sin38° × Sin52° = 0


LHS = Cos38° × Cos52° - Sin38° × Sin52°




=> Cos38° × Cos( 90 - 38 ) - Sin38° × Sin(90-38° ).




=> Cos38° × Sin38° - Sin38° × Cos38°





=> 0


RHS = 0


Hence,

LHS = RHS = 0