Question

Prove that $\frac{sin 150\textdegree - 5 cos 300\textdegree + 7 tan 225\textdegree}{tan 135\textdegree + 3 sin 210\textdegree}$ = -2.

Answer 1

we know, sin(π - x) = sinx
so, sin150° = sin(180° - 30°) = sin30° = 1/2

cos(2π - x) = cosx
so, cos300° = cos(360° - 60°) = cos60° = 1/2

tan(π + x) = tanx
so, tan225° = tan(180° + 45°) = tan45° = 1

tan(π - x) = -tanx
so, tan135° = tan(180° - 45°) = -tan45° = -1

sin(π + x) = -sinx
so, sin210° = sin(180° + 30°) = -sin30° = -1/2

now, LHS = (sin150° - 5cos300° + 7tan225°)/(tan135° + 3sin210° )

= (1/2 - 5 × 1/2 + 7)/(-1 + 3 × -1/2)

= (-2 + 7)/(-5/2)

= 5/(-5/2)

= -2 = RHS

Answer 2

HELLO DEAR,



we know:-
sin(π - x) = sinx
so, sin150° = sin(180° - 30°) = sin30° = 1/2
cos(2π - x) = cosx
so, cos300° = cos(360° - 60°) = cos60° = 1/2
tan(π + x) = tanx
so, tan225° = tan(180° + 45°) = tan45° = 1
tan(π - x) = -tanx
so, tan135° = tan(180° - 45°) = -tan45° = -1
sin(π + x) = -sinx
so, sin210° = sin(180° + 30°) = -sin30° = -1/2

now,
{sin150° - 5cos300° + 7tan225°}/{tan135° + 3sin210°}

=> (1/2 - 5 × 1/2 + 7)/(-1 + 3 × -1/2)

=> (-2 + 7)/(-5/2)

=> 5/(-5/2)

=> -2



I HOPE IT'S HELP YOU DEAR,
THANKS