Question

Evaluate 2sin68 / cos22 - 2tan(90-15)/5cot15- 3tan45 tan20 tan40 tan50 tan70 /5(sin square70+sin square20).

Answer 1

sin68° = sin(90°—22°) = cos22°
sin68°÷cos22° = 1

tan(90°—15°) = cot15°
tan(90°—15°)÷cot15° = 1

tan20° = tan(90°—70°) = cot70° = 1÷tan70°
tan20° × tan70° = 1

tan40° = tan(90°—50°) = cot50° = 1÷tan50°
tan40° × tan50° = 1

sin²70° + sin²20° = sin²(90°—20°) + sin²20°
= cos²20° + sin²20°
= 1

2×(sin68°÷cos22°) — (2÷5){tan(90°—15°)÷cot15°} —(3 tan45° tan20° tan 40° tan50° tan70°) ÷ {5(sin²70°+sin²20°)}
=2×1 — (2÷5)×1 —(3tan45°×1×1)÷(5×1)
=2 — (2÷5) — (3÷5)
=2—1
=1

Answer 2

Answer:

LHS = 1

Step-by-step explanation:

ATQ, We have to find;

$\sf \Longrightarrow \dfrac{2 \ sin68^\circ}{cos22^\circ} - \dfrac{2 \ tan(90^\circ-15^\circ)}{5 \ cot15^\circ} - \dfrac{3tan45^\circ \ tan20^\circ \ tan40^\circ \ tan50^\circ \ tan70^\circ}{5(sin^{2}70^\circ + sin^{2}20)}$

Using,

⇒ tan(90° - 15°) = cot15°

⇒ cos22° = cos(90° - 68°)

⇒ tan45° = 1

$\sf \Longrightarrow \dfrac{2 \ sin68^\circ}{cos(90^\circ - 68^\circ)} - \dfrac{2 \ cot15^\circ}{5 \ cot15^\circ} - \dfrac{3(1) \ tan20^\circ \ tan40^\circ \ tan50^\circ \ tan70^\circ}{5(sin^{2}70^\circ + sin^{2}20)}$

Using,

⇒ cos(90° - 68°) = sin68°

⇒ tan20° = tan(90° - 70°)

⇒ tan40° = tan(90° - 50°)

$\sf \Longrightarrow \dfrac{2 \ sin68^\circ}{sin \ 68^\circ} - \dfrac{2}{5} - \dfrac{3 \times tan(90^\circ - 70^\circ) \ tan(90^\circ - 50^\circ) \ tan50^\circ \ tan70^\circ}{5(sin^{2}70^\circ + sin^{2}20)}$

Using,

⇒ tan(90° - 70°) = cot70°

⇒ tan(90° - 50°) = cot50°

$\sf \Longrightarrow \dfrac{2}{1} - \dfrac{2}{5} - \dfrac{3 \times cot70^\circ \ cot50^\circ \ tan50^\circ \ tan70^\circ}{5(sin^{2}70^\circ + sin^{2}20)}$

$\sf \Longrightarrow \dfrac{2}{1} - \dfrac{2}{5} - \dfrac{3 \times cot70^\circ \ tan70^\circ \ cot50^\circ \ tan50^\circ}{5(sin^{2}70^\circ + sin^{2}20)}$

Using,

⇒ tan70° × cot70° = 1

⇒ tan50° × cot50° = 1

⇒ sin²20° = sin²(90° - 70°)

$\sf \Longrightarrow \dfrac{2}{1} - \dfrac{2}{5} - \dfrac{3 \times 1 \times 1}{5(sin^{2}70^\circ + sin^{2}(90^\circ - 70^\circ)}$

⇒ Using sin²(90° - 70°) = cos²70°

$\sf \Longrightarrow \dfrac{2}{1} - \dfrac{2}{5} - \dfrac{3}{5(sin^{2}70^\circ + cos^{2}70^\circ)}$

⇒ Using sin²70° + cos²70° = 1

$\sf \Longrightarrow \dfrac{2}{1} - \dfrac{2}{5} - \dfrac{3}{5(1)}$

$\sf \Longrightarrow \dfrac{2}{1} - \dfrac{2}{5} - \dfrac{3}{5}$

$\sf \Longrightarrow \dfrac{10 - 3 - 2}{5}$

$\sf \Longrightarrow \dfrac{7 - 2}{5}$

$\sf \Longrightarrow \dfrac{5}{5}$

$\sf \Longrightarrow 1$