Trigonometry ratios practice · Medium

Option C ($2m$) is correct because simplifying the given expression using trigonometric identities leads to the result $2m$.

Option A ($\left( {0,\;\pi /2} \right)$) is correct because the function changes sign within this interval, indicating the presence of a root.

Option B ($\frac{{3}}{{4}} \leq A \leq 1$) is correct because the expression can be rewritten in terms of $\sin^2 2\theta$ and its range can be determined.

Option A ($x > 0$) is correct because $\sin A$ is positive and $\cos A$ is negative, but $\sin A$ has a larger magnitude, resulting in a positive sum.

Option A is correct because it derives the relationship between tan α tan β, x, y, a, and b based on the given equations and trigonometric identities.

Option B ($-\frac{{2}}{{\sin \alpha}}$) is correct because, within the given range for alpha, the expression simplifies to this value considering the sign of sine in that quadrant.

Option B ($y \notin [1/3, 3]$) is correct because solving for $y = \frac{{\tan x}}{{\tan 3x}}$ leads to the conclusion that y cannot lie within the closed interval [1/3, 3].

Option B ($a^2+b^2-3$) is correct because by using the given relations and trigonometric identities, the expression $\frac{\cos 3 \theta}{\cos \theta}$ simplifies to $a^2+b^2-3$.

Option C ($\cos 1$) is correct because the cosine function is decreasing on the interval [0, 1], and since $\cos x$ varies from -1 to 1, $\cos(\cos x)$ is minimized when $\cos x$ is closest to 1 (in magnitude but negative), which is when $\cos x = 1$ or $\cos x = -1$, so the minimum value is $\cos(1)$.

Option A (5) is correct because by using trigonometric identities and the given value of $\tan 6\theta$, the expression simplifies to 5.

Option B (4) is correct because solving the equation derived from the sum of angles in an A.P. equal to the sum of interior angles of a polygon yields n=4.

Option C ($9 \frac{1}{2}$) is correct because pairing $\sin^2 x$ with $\sin^2 (90-x) = \cos^2 x$ allows us to use the identity $\sin^2 x + \cos^2 x = 1$, and accounting for the remaining terms.

Option B ($8 \cos 2 A$) is correct because the expression simplifies to $8 \cos(2A)$ using trigonometric identities.

Option B (2) is correct because by expanding the determinant and using the trigonometric identity $\tan A + \tan B + \tan C = \tan A \tan B \tan C$ for a triangle, the determinant simplifies to 2.

Option A (1) is correct because the given condition $\sin x + \sin^2 x = 1$ leads to $\sin x = \cos^2 x$, which simplifies the expression to 1.

Option B ($\frac{\sqrt{10}-10}{12}$) is correct because it is the value of tan α obtained by solving the trigonometric equation and considering the given range for α.

Option C ($\sqrt{3} / 2$) is correct because the expression is a direct application of the cosine double angle formula: $\cos(2\theta) = \frac{{1 - \tan^2(\theta)}}{{1 + \tan^2(\theta)}}$ with $\theta = 15^{\circ}$.

Option A ($\frac{p^2}{\{p^2+(1-q)^2\}}$) is correct because by using the relationship between the roots and coefficients of the quadratic equation and the tangent addition formula, the expression simplifies to the given form.

Option B is correct because expanding and simplifying the expression leads to 2(1 - sin A)(1 + cos A).

Option A ($m\,\,{\left( {m{n^2}} \right)^{1/3}} - n\,\,{\left( {n{m^2}} \right)^{1/3}} = 1$) is correct because it is the simplified form of the equation after substituting and using trigonometric identities.

Option B ($- \,\sqrt 7 \,\sin \,A$) is correct because it is derived by substituting A with 90° + A in the given equation.

Option A (1) is correct because the given condition leads to $\sin x = \cos^2 x$, and substituting this into the expression simplifies it to 1.

Option A (12) is correct because substituting A=B=C=60 degrees into the given equation and simplifying leads to k=12.

Option B (isosceles) is correct because the given condition leads to the conclusion that two angles of the triangle are equal, making it isosceles.

Option A ($\mathrm{a}=\sqrt{{2}} \mathrm{c}$) is correct because equating the perpendicular distance from the center of the circle to the tangent and the y-intercept of the tangent to the parabola gives this relationship.

Option A ($\cot \alpha \cot \beta \cot \gamma$) is correct because using the identity $sec^2 x - tan^2 x = 1$ and the given equation, we can derive the expression.

Option B (-1/2) is correct because the sum of cosines can be simplified using trigonometric identities to obtain the result -1/2.

Option A (6) is correct because by substituting the known values of $\cos 36^{\circ}$ and $\cot (\frac{15}{2})^{\circ}$, the equation simplifies to $\sum_{r=1}^6 \sqrt{r}$, thus k=6.

Option A (1) is correct because the expression simplifies to 1 using the trigonometric identity (1+tanA)(1+tanB)=2 when A+B=45 degrees.

Option B ($\sec A - \tan A$) is correct because simplifying the given expression using trigonometric identities leads to this result, considering the given range of A.