Given:\( \cos 9 \alpha=\sin \alpha \) and \( 9 \alpha

Given:\( \left(\tan 1^{\circ} \tan 2^{\circ} \tan 3^{\circ} \ldots \tan 89^{\circ}\right) \)To do:We have to find the value of \( \left(\tan 1^{\circ} \tan 2^{\circ} \tan 3^{\circ} \ldots \tan 89^{\circ}\right) \).Solution:  We know that, $tan\ (90^{\circ}- \theta) = cot\ \theta$$tan\ \theta \times \cot\ \theta=1$Therefore, $\tan 1^{\circ} \tan 2^{\circ} \tan 3^{\circ}.....\tan 45^{\circ}...... \tan 87^{\circ} \tan 88^{\circ} \tan 89^{\circ}=\tan 1^{\circ} \tan 2^{\circ} \tan 3^{\circ}......\tan 45^{\circ}.........\tan (90^{\circ}-3^{\circ}) \tan (90^{\circ}-2^{\circ}) \tan (90^{\circ}-1^{\circ})$$=\tan 1^{\circ} \tan 2^{\circ} \tan 3^{\circ}........\tan 45^{\circ}......\cot 3^{\circ} \cot 2^{\circ} \cot 1^{\circ}$$=(\tan 1^{\circ} \cot 1^{\circ})(\tan 2^{\circ} \cot 2^{\circ})..................(\tan 44^{\circ} \cot 44^{\circ})(1)$$=1$Read More

Given:\( \cos (\alpha+\beta)=0 \)To do:We have to find the value of \( \sin (\alpha-\beta) \)Solution:  $\cos (\alpha+\beta) =0$$=\cos 90^{\circ}$          [Since $\cos 90^{\circ}=0$]This implies,$\alpha+\beta =90^{\circ}$$\alpha =90^{\circ}-\beta$$\sin\ (\alpha-\beta) =\sin (90^{\circ}-\beta-\beta)$$=\sin\ (90^{\circ}-2 \beta)$$=\cos 2 \beta$           [Since $\sin\ (90^{\circ}-\theta)=\cos \theta$]Hence, $\sin (\alpha-\beta)$ can be reduced to $\cos 2 \beta$.

Given:$sin\ \theta = \frac{a}{b}$To do:We have to find the value of $cos\ \theta$.Solution:  We know that,$sin^2 \theta+cos^2 \theta=1$$cos \theta=\sqrt{1-sin^2 \theta}$Therefore,$cos \theta= \sqrt{1-(\frac{a}{b})^2}$$=\sqrt{1-\frac{a^2}{b^2}}$$=\sqrt{\frac{b^2-a^2}{b^2}}$$=\frac{\sqrt{b^2-a^2}}{b}$

Given:\( \left[\operatorname{cosec}\left(75^{\circ}+\theta\right)-\sec \left(15^{\circ}-\theta\right)-\tan \left(55^{\circ}+\theta\right)+\right. \) \( \left.\cot \left(35^{\circ}-\theta\right)\right] \)To do:We have to evaluate \( \left[\operatorname{cosec}\left(75^{\circ}+\theta\right)-\sec \left(15^{\circ}-\theta\right)-\tan \left(55^{\circ}+\theta\right)+\right. \) \( \left.\cot \left(35^{\circ}-\theta\right)\right] \).Solution:  We know that, $\operatorname{cosec}\ (90^{\circ}- \theta) =\sec\ \theta$$cot\ (90^{\circ}- \theta) = tan\ \theta$Therefore, $\operatorname{cosec}\left(75^{\circ}+\theta\right)-\sec \left(15^{\circ}-\theta\right)-\tan \left(55^{\circ}+\theta\right)+\cot \left(35^{\circ}-\theta\right)$$=\operatorname{cosec}(90^{\circ}-(15^{\circ}-\theta))-\sec \left(15^{\circ}-\theta\right)-\tan \left(55^{\circ}+\theta\right)+\cot (90^{\circ}-(55^{\circ}+\theta))$$=\sec (15^{\circ}-\theta)-\sec \left(15^{\circ}-\theta\right)-\tan \left(55^{\circ}+\theta\right)+\tan (55^{\circ}+\theta)$$=0$Read More

Given:$sin\ A = \frac{1}{2}$To do:We have to find the value of $cot\ A$.Solution:  Let, in a triangle $ABC$ right-angled at $B$, $sin\ A=\frac{1}{2}$.We know that, In a right-angled triangle $ABC$ with right angle at $B$, By Pythagoras theorem, $AC^2=AB^2+BC^2$By trigonometric ratios definitions, $sin\ \theta=\frac{Opposite}{Hypotenuse}=\frac{BC}{AC}$$cos\ \theta=\frac{Adjacent}{Hypotenuse}=\frac{AB}{AC}$$sec\ \theta=\frac{Hypotenuse}{Adjacent}=\frac{AC}{AB}$$tan\ \theta=\frac{Opposite}{Adjacent}=\frac{BC}{AB}$$cot\ \theta=\frac{Adjacent}{Opposite}=\frac{AB}{BC}$Here, $AC^2=AB^2+BC^2$$\Rightarrow (2)^2=(AB)^2+1^2$$\Rightarrow AB^2=4-1$$\Rightarrow AB=\sqrt{3}$Therefore, $cot\ \A=\frac{AB}{BC}=\frac{\sqrt3}{1}$$=\sqrt3$. Read More

Given:$cos\ A = \frac{4}{5}$ To do:We have to find the value of $tan\ A$.Solution:  Let, in a triangle $ABC$ right-angled at $B$, $cos\ A=\frac{4}{5}$.We know that, In a right-angled triangle $ABC$ with right angle at $B$, By Pythagoras theorem, $AC^2=AB^2+BC^2$By trigonometric ratios definitions, $sin\ \theta=\frac{Opposite}{Hypotenuse}=\frac{BC}{AC}$$cos\ \theta=\frac{Adjacent}{Hypotenuse}=\frac{AB}{AC}$$sec\ \theta=\frac{Hypotenuse}{Adjacent}=\frac{AC}{AB}$$tan\ \theta=\frac{Opposite}{Adjacent}=\frac{BC}{AB}$Here, $AC^2=AB^2+BC^2$$\Rightarrow (5)^2=(4)^2+BC^2$$\Rightarrow BC^2=25-16$$\Rightarrow BC=\sqrt{9}=3$Therefore, $tan\ A=\frac{BC}{AB}=\frac{3}{4}$. Read More

Given:Students of a school are standing in rows and columns in their playground for a drill practice. \( \mathrm{A}, \mathrm{B}, \mathrm{C} \) and \( \mathrm{D} \) are the positions of four students.To do:We have to find whether it is possible to place Jaspal in the drill in such a way that he is equidistant from each of the four students \( A, B, C \) and D.Solution:From the figure, we observe that the positions of the four students A, B, C and D are $(3, 5), (7, 9), (11, 5)$ and $(7, 1)$ respectively.The four vertices form a quadrilateral. $A B ... Read More

Given:The points $A (1, -2), B (2, 3), C (a, 2)$ and $D (-4, -3)$ form a parallelogram.To do:We have to find the value of $a$ and height of the parallelogram taking $AB$ as base.Solution:Draw a perpendicular from \( \mathrm{D} \) to \( \mathrm{AB} \) which meets \( \mathrm{AB} \) at \( \mathrm{P} \).\( \mathrm{DP} \) is the height of the parallelogram. We know that,  Diagonals of a parallelogram bisect each other.This implies, Mid-point of $AC =$ Mid-point of $BD$The mid-point of a line segment joining points \( \left(x_{1}, y_{1}\right) \) and \( \left(x_{2}, y_{2}\right) \) is \( (\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}) \)\( (\frac{1+a}{2}, ... Read More

Given:The points \( A\left(x_{1}, y_{1}\right), \mathrm{B}\left(x_{2}, y_{2}\right) \) and \( \mathrm{C}\left(x_{3}, y_{3}\right) \) are the vertices of \( \Delta \mathrm{ABC} \)To do:We have to find the coordinates of the centroid of the triangle ABC.Solution:We know that,Coordinates of the centroid of a triangle $=\left(\frac{\text { Sum of abscissa of all vertices, }}{3}, \frac{\text { Sum of ordinate of all vertices }}{3}\right)$Therefore,The coordinates of the centroid of the triangle ABC $=\left(\frac{x_{1}+x_{2}+x_{3}}{3}, \frac{y_{1}+y_{2}+y_{3}}{3}\right)$