Question

Trigeonometric Identities

Prove
(1 - tan²theta) / (cos²theta - 1) = tan²theta

If you don't know the answer then don't put rubbish answers here then account will be banned permanently

Answer 1

Correct Question :-

Prove that,

( 1 - tan²Φ)/( cot²Φ - 1 ) = tan²Φ

Solution :-

( 1 - tan²Φ ) / ( cot²Φ - 1 ) = tan²Φ

By taking LHS :-

[ As, we know that, cot²Φ = 1 / tan²Φ]

( 1 - tan²Φ )/( 1/tan²Φ - 1/1) = 0

By taking LCM, we get :-

( 1 - tan²Φ ) / (1 - tan²Φ/tan²Φ) = tan²Φ

( 1 - tan²Φ ) × tan²Φ/( 1 - tan²Φ)= tan²Φ

tan²Φ = tan²Φ

Hence, Proved

Some Trigonometric identities :-

• Sin²Φ + Cos²Φ = 1

• Sec²Φ - tan²Φ = 1

• cosec²Φ - cot²Φ = 1

Trigonometric ratios :-

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

Answer 2

Answer:

In aqueous solution, precipitation is the process of transforming a dissolved substance into an insoluble solid from a super-saturated solution. The solid formed is called the precipitate.

The *ucking moderators are deleting my ans. que.

so I am going to my ItzHeartHackerKing I'd see in my following na