We use the identity:
\[ \tan(90^\circ - x) = \cot x. \]Rewriting the given expression:
\[ \tan 9^\circ - \tan 27^\circ - \tan 63^\circ + \tan 81^\circ. \]Using symmetry properties:
\[ \tan 81^\circ = \cot 9^\circ, \quad \tan 63^\circ = \cot 27^\circ. \]Thus, the given expression transforms into:
\[ \tan 9^\circ - \tan 27^\circ - \cot 27^\circ + \cot 9^\circ. \]Using the identity:
\[ \tan x - \cot x = \frac{2 \tan 2x}{1 - \tan^2 x}, \]we simplify and evaluate:
\[ 4. \]Final Answer: \( \mathbf{4} \).
The area enclosed by the closed curve $C$ given by the differential equation $\frac{d y}{d x}+\frac{x+a}{y-2}=0, y(1)=0$ is $4 \pi$.
Let $P$ and $Q$ be the points of intersection of the curve $C$ and the $y$-axis If normals at $P$ and $Q$ on the curve $C$ intersect $x$-axis at points $R$ and $S$ respectively, then the length of the line segment $R S$ is
The statement
\((p⇒q)∨(p⇒r) \)
is NOT equivalent to
Let α, β(α > β) be the roots of the quadratic equation x2 – x – 4 = 0.
If \(P_n=α^n–β^n, n∈N\) then \(\frac{P_{15}P_{16}–P_{14}P_{16}–P_{15}^2+P_{14}P_{15}}{P_{13}P_{14}}\)
is equal to _______.
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)