Question

If 25 sin Q= 7 , find 2tanQ/1+tan^2Q​

Answer 1

EXPLANATION.

⇒ 25 sin(Q) = 7.

As we know that,

sinθ = Perpendicular/Hypotenuse.

⇒ sin(Q) = 7/25 = P/H.

By using Pythagoras Theorem, we get.

⇒ H² = P² + B².

Hypotenuse > Perpendicular > Base.

Using this formula in the equation, we get.

⇒ (25)² = (7)² + B².

⇒ 625 = 49 + B².

⇒ 625 - 49 = B².

⇒ 576 = B².

⇒ B = √576.

⇒ B = 24.

tanθ = Perpendicular/Base = 7/24.

To find :

⇒ 2tan(Q)/1 + tan²(Q).

⇒ [2 x (7/24)/1 + (7/24)²].

⇒ [(14/24)/1 + 49/576].

⇒ [(14/24)/576 + 49/576].

⇒ [(14/24)/625/576].

⇒ [14/24 x 576/625].

⇒ 336/625.

⇒ 2tan(Q)/1 + tan²(Q) = 336/625.

Answer 2

Question

If 25 sin Q= 7 , find 2tanQ/1+tan^2Q

Answer

➽ sec∅ = 25/7 = H/B

➽ H²=P²+B²

➽ (25)²= P²+(7)²

➽625 = P²+(49)

➽ 576 = P²

➽P = 24 cm

Tan ∅ = P/B = 27/7

To find

➽ 2 tan (Q)/1+tan²(Q)

[2×(7×24)/1+(7×24)²]

[(14/24)/1+49/576]

[(14/24)/576+49/576]

[(14/24)/625/576]

[(14/24×576/625]

$\to \tt \: \frac{336}{625}$