Question

it Sin3A=1, what is the value of tan²A-Sec²A?​

Answer 1

Answer:

-1

Step-by-step explanation:

There's an identity in Trigonometry:

Sec²A = 1 + Tan²A

Tan²A - Sec²A = -1

The angle and the given equation (Sin3A = 1) doesn't matter

Answer 2

ㅤ✠Given :-

sin3A = 1

ㅤ✠To find :-

tan²A- sec²A

ㅤ✠Solution :-

As they given,

sin3A = 1

1 can be written as sin90° So,

sin3A = sin90°

ㅤ✠Removing "sin" on both sides

3A = 90°

A = 90°/3

A = 30°

So,

tan²A - sec²A = tan²30° - sec²30°

⟹tan30° = 1/√3

⟹sec30° = 2/√3

ㅤ✠Substituting the values,

⟹ (1/√3)² - (2/√3)²

⟹ 1/3 - 4/3

⟹ (1-4)/3

⟹ -3/3

⟹ -1

ㅤSo, the value of tan²A - sec²A is -1

ㅤㅤㅤ______________

✠Know more :-

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0\end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$