Prove the following identities (1-16)
Question: Prove the following identities (1-16) $1-\frac{\sin ^{2} x}{1+\cot x}-\frac{\cos ^{2} x}{1+\tan x}=\sin x \cos x$ Solution: $1-\frac{\sin ^{2} x}{1+\cot x}-\frac{\cos ^{2} x}{1+\tan x}=\sin x \cos x$ $\mathrm{LHS}=1-\frac{\sin ^{3} x}{\sin x+\cos x}-\frac{\cos ^{3} x}{\sin x+\cos x}$ $=\frac{\sin x+\cos x-\left(\sin ^{3} x+\cos ^{3} x\right)}{\sin x+\cos x}$ $=\frac{(\sin x+\cos x)\left(1-\sin ^{2} x-\cos ^{2} x+\sin x \cos x\right)}{\sin x+\cos x}$ $=\left(1-\sin ^{2} x-\cos ^{2} x+\s...
Read More →In the given figure, D is the mid-point of side BC and AE ⊥ BC
Question: In the given figure, $\mathrm{D}$ is the mid-point of side $\mathrm{BC}$ and $\mathrm{AE} \perp \mathrm{BC}$. If $\mathrm{BC}=\mathrm{a}, \mathrm{AC}=b, \mathrm{AB}=c, \mathrm{ED}=x, \mathrm{AD}=p$ and $\mathrm{AE}=h$, prove that : (i) $b^{2}=p^{2}+a x+\frac{a^{2}}{4}$ (ii) $c^{2}=p^{2}-a x+\frac{a^{2}}{4}$ (iii) $b^{2}+c^{2}=2 p^{2}+\frac{a^{2}}{2}$ Solution: (i) It is given that $\mathrm{D}$ is the midpoint of $\mathrm{BC}$ and $B C=a$. Therefore, $B D=D C=\frac{a}{2}$...(1) Using Py...
Read More →Prove the following identities (1-16)
Question: Prove the following identities (1-16) Solution: $\mathrm{LHS}=\frac{\tan ^{3} x}{1+\tan ^{2} x}+\frac{\cot ^{3} x}{1+\cot ^{2} x}$ $=\frac{\tan ^{3} x}{\sec ^{2} x}+\frac{\cot ^{3} x}{\operatorname{cosec}^{2} x}$ $=\frac{\frac{\sin ^{3} x}{\cos ^{3} x}}{\frac{1}{\cos ^{2} x}}+\frac{\frac{\cos ^{3} x}{\sin ^{3} x}}{\frac{1}{\sin ^{2} x}}$ $=\frac{\sin ^{3} x}{\cos ^{3} x} \times \frac{\cos ^{2} x}{1}+\frac{\cos ^{3} x}{\sin ^{3} x} \times \frac{\sin ^{2} x}{1}$ $=\frac{\sin ^{3} x}{\cos...
Read More →Find the area of the region bounded by the ellipse
Question: Find the area of the region bounded by the ellipse $\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$ Solution: The given equation of the ellipse can be represented as $\frac{x^{2}}{4}+\frac{y^{2}}{9}=1$ $\Rightarrow y=3 \sqrt{1-\frac{x^{2}}{4}}$ ...(1) It can be observed that the ellipse is symmetrical aboutx-axis andy-axis. Area bounded by ellipse = 4 Area OAB $\therefore$ Area of $\mathrm{OAB}=\int_{0}^{2} y d x$ $=\int_{0}^{2} 3 \sqrt{1-\frac{x^{2}}{4}} d x \quad[$ Using (1) $]$ $=\frac{3}{2} \in...
Read More →Prove the following identities (1-16)
Question: Prove the following identities (1-16) $\frac{\cos x}{1-\sin x}=\frac{1+\cos x+\sin x}{1+\cos x-\sin x}$ Solution: RHS $=\frac{1+\cos x+\sin x}{1+\cos x-\sin x}$ $=\frac{(1+\cos \mathrm{x})+(\sin \mathrm{x})}{(1+\cos \mathrm{x})-(\sin \mathrm{x})}$ $=\frac{[(1+\cos \mathrm{x})+(\sin \mathrm{x})][(1+\cos \mathrm{x})+(\sin \mathrm{x})]}{[(1+\cos \mathrm{x})-(\sin \mathrm{x})][(1+\cos \mathrm{x})+(\sin \mathrm{x})]}$ $=\frac{[(1+\cos x)+(\sin x)]^{2}}{(1+\cos x)^{2}-(\sin x)^{2}}$ $=\frac{...
Read More →In figure, O is the centre of the circle. Find ∠BAC.
Question: In figure, O is the centre of the circle. Find BAC. Solution: We haveAOB = 80 AndAOC = 110 Therefore,AOB + AOC + BOC = 360 [Complete angle] ⇒ 80 + 100 + BOC = 360 ⇒ BOC = 360 80 110 ⇒ BOC = 170 By degree measure theorem BOC = 2BAC ⇒ 170 = 2BAC ⇒ BAC = 170/2 = 85...
Read More →Find the area of the region bounded by the ellipse
Question: Find the area of the region bounded by the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$ Solution: The given equation of the ellipse, $\frac{x^{2}}{16}+\frac{y^{2}}{9}=1$, can be represented as It can be observed that the ellipse is symmetrical about $x$-axis and $y$-axis. $\therefore$ Area bounded by ellipse $=4 \times$ Area of $\mathrm{OAB}$ Area of $\mathrm{OAB}=\int_{0}^{4} y d x$ $=\int_{0}^{4} 3 \sqrt{1-\frac{x^{2}}{16}} d x$ $=\frac{3}{4} \int_{0}^{1} \sqrt{16-x^{2}} d x$ $=\frac...
Read More →In figure, it is given that O is the centre of the circle and
Question: In figure, it is given that O is the centre of the circle and AOC = 150. Find ABC. Solution: AOC = 150 AOC + reflex AOC = 360 [Complex angle] ⇒ 150 + reflex AOC = 360 ⇒ reflex AOC = 360 150 ⇒ reflex AOC = 210 ⇒ 2ABC = 210 [By degree measure theorem] ⇒ ABC = 210/2 = 105...
Read More →In figure, O is the centre of the circle.
Question: In figure, O is the centre of the circle. If APB = 50, find AOB and OAB. Solution: APB = 50 By degree measure theorem AOB = 2APB ⇒ APB = 2 50 = 100 since OA = OB [Radius of circle] Then OAB = OBA [Angles opposite toequalsides] Let OAB = x In ΔOAB, by angle sum property OAB + OBA + AOB = 180 ⟹ x + x + 100= 180 ⟹ 2x = 180- 100 ⟹ 2x = 80 ⟹ x = 40 OAB = OBA = 40...
Read More →Find the area of the region bounded by
Question: Find the area of the region bounded by $x^{2}=4 y, y=2, y=4$ and the $y$-axis in the first quadrant. Solution: The area of the region bounded by the curve, $x^{2}=4 y, y=2$, and $y=4$, and the $y$-axis is the area $\mathrm{ABCD}$. Area of $\mathrm{ABCD}=\int_{2}^{4} x d y$ $=\int_{2}^{4} 2 \sqrt{y} d y$ $=2 \int_{2}^{4} \sqrt{y} d y$ $=2\left[\frac{y^{\frac{3}{2}}}{\frac{3}{2}}\right]_{2}^{4}$ $=\frac{4}{3}\left[(4)^{\frac{3}{2}}-(2)^{\frac{3}{2}}\right]$ $=\frac{4}{3}[8-2 \sqrt{2}]$ $...
Read More →A circular park of radius 40 m is situated in a colony. Three boys Ankur,
Question: A circular park of radius 40 m is situated in a colony. Three boys Ankur, Amit and Anand are sitting at equal distance on its boundary each having a toy telephone in his hands to talk to each other. Find the length of the string of each phone. Solution: Given that AB = BC = CA So, ABC is an equilateral triangle OA (radius) = 40 m Medians of equilateral triangle pass through the circum centre (O) of the equilateral triangle ABC. We also know that median intersect each other at the ratio...
Read More →Find the area of the region bounded by
Question: Find the area of the region bounded by $y^{2}=9 x, x=2, x=4$ and the $x$-axis in the first quadrant. Solution: The area of the region bounded by the curve, $y^{2}=9 x, x=2$, and $x=4$, and the $x$-axis is the area ABCD. Area of $\mathrm{ABCD}=\int_{2}^{1} y d x$ $=\int_{2}^{4} 3 \sqrt{x} d x$ $=3\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_{2}^{4}$ $=2\left[x^{\frac{3}{2}}\right]_{2}^{4}$ $=2\left[(4)^{\frac{3}{2}}-(2)^{\frac{3}{2}}\right]$ $=2[8-2 \sqrt{2}]$ $=(16-4 \sqrt{2})$ uni...
Read More →Prove the following identities (1-16)
Question: Prove the following identities (1-16) $(\sec x \sec y+\tan x \tan y)^{2}-(\sec x \tan y+\tan x \sec y)^{2}=1$ Solution: $\mathrm{LHS}=(\sec x \sec y+\tan x \tan y)^{2}-(\sec x \tan y+\tan x \sec y)^{2}$ $=\left[(\sec x \sec y)^{2}+(\tan x \tan y)^{2}-2(\sec x \sec y)(\tan x \tan y)\right]$ $-\left[(\sec x \tan y)^{2}+(\tan x \sec y)^{2}-2(\sec x \tan y)(\tan x \sec y)\right]$ $=\left[\begin{array}{llll}\sec ^{2} x \sec ^{2} y+\tan ^{2} x \tan ^{2} y-2 \sec x \sec y \tan x \tan y\end{ar...
Read More →Prove the following identities (1-16)
Question: Prove the following identities (1-16) $(\sec x \sec y+\tan x \tan y)^{2}-(\sec x \tan y+\tan x \sec y)^{2}=1$ Solution: $\mathrm{LHS}=(\sec x \sec y+\tan x \tan y)^{2}-(\sec x \tan y+\tan x \sec y)^{2}$ $=\left[(\sec x \sec y)^{2}+(\tan x \tan y)^{2}-2(\sec x \sec y)(\tan x \tan y)\right]$ $-\left[(\sec x \tan y)^{2}+(\tan x \sec y)^{2}-2(\sec x \tan y)(\tan x \sec y)\right]$ $=\left[\begin{array}{llll}\sec ^{2} x \sec ^{2} y+\tan ^{2} x \tan ^{2} y-2 \sec x \sec y \tan x \tan y\end{ar...
Read More →Find the area of the region bounded by the curve
Question: Find the area of the region bounded by the curve $y^{2}=x$ and the lines $x=1, x=4$ and the $x$-axis. Solution: The area of the region bounded by the curve, $y^{2}=x$, the lines, $x=1$ and $x=4$, and the $x$-axis is the area $A B C D$. Area of $\mathrm{ABCD}=\int_{1}^{4} y d x$ $=\int_{1}^{4} \sqrt{x} d x$ $=\left[\frac{x^{\frac{3}{2}}}{\frac{3}{2}}\right]_{1}^{4}$ $=\frac{2}{3}\left[(4)^{\frac{3}{2}}-(1)^{\frac{3}{2}}\right]$ $=\frac{2}{3}[8-1]$ $=\frac{14}{3}$ units...
Read More →Three girls Ishita, Isha and Nisha are playing a game by standing on a circle of radius 20 m drawn in a park.
Question: Three girls Ishita, Isha and Nisha are playing a game by standing on a circle of radius 20 m drawn in a park. Ishita throws a ball to Isha, Isha to Nisha and Nisha to Ishita. If the distance between Ishita and Isha and between Isha and Nisha is 24m each, what is the distance between Ishita and Nisha. Solution: Let R, S and M be the position of Ishita, Isha and Nisha respectively. AR = AS = 24/2 = 12 cm OR = OS = OM = 20 cm [Radii of circle] In ΔOAR, $O A^{2}+A R^{2}=O R^{2}$ $\mathrm{O...
Read More →Prove the following identities (1-16)
Question: Prove the following identities (1-16) $\frac{\sin ^{3} x+\cos ^{3} x}{\sin x+\cos x}+\frac{\sin ^{3} x-\cos ^{3} x}{\sin x-\cos x}=2$ Solution: $\mathrm{LHS}=\frac{\sin ^{3} x+\cos ^{3} x}{\sin x+\cos x}+\frac{\sin ^{3} x-\cos ^{3} x}{\sin x-\cos x}$ $=\frac{(\sin x+\cos x)\left[\sin ^{2} x+\cos ^{2} x-\sin x \cos x\right]}{(\sin x+\cos x)}+\frac{(\sin x-\cos x)\left[\sin ^{2} x+\cos ^{2} x+\sin x \cos x\right]}{(\sin x-\cos x)}$ $=\left[\sin ^{2} x+\cos ^{2} x-\sin x \cos x\right]+\le...
Read More →The lengths of two parallel chords of a circle are 6 cm and 8 cm.
Question: The lengths of two parallel chords of a circle are 6 cm and 8 cm. If the smaller chord is at a distance of 4 cm from the centre, what is the distance of the other chord from the centre? Solution: Distance of smaller chord AB from centre of circle = 4 cm, OM = 4 cm MB = AB/2 = 6/2 = 3 cm In ΔOMB $\mathrm{OM}^{2}+\mathrm{MB}^{2}=\mathrm{OB}^{2}$ $4^{2}+9^{2}=\mathrm{OB}^{2}$ $16+9=\mathrm{OB}^{2}$ $O B=\sqrt{25}$ OB = 5 cm In ΔOND OD = OB = 5 cm [Radii of same circle] ND = CD/2 = 8/2 = 4...
Read More →Prove the following identities (1-16)
Question: Prove the following identities (1-16) $\frac{\tan x}{1-\cot x}+\frac{\cot x}{1-\tan x}=(\sec x \operatorname{cossec} x+1)$ Solution: $\mathrm{LHS}=\frac{\tan x}{1-\cot x}+\frac{\cot x}{1-\tan x}$ $=\frac{\frac{\sin x}{\cos x}}{1-\frac{\cos x}{\sin x}}+\frac{\frac{\cos x}{\sin x}}{1-\frac{\sin x}{\cos x}}$ $=\frac{\frac{\sin x}{\cos x}}{\frac{\sin x-\cos x}{\sin x}}+\frac{\frac{\cos x}{\sin x}}{\frac{\cos x-\sin x}{\cos x}}$ $=\frac{\sin x}{\cos x} \times \frac{\sin x}{\sin x-\cos x}+\f...
Read More →In right-angled triangle ABC is which ∠C = 90°,
Question: In right-angled triangle $\mathrm{ABC}$ is which $\angle \mathrm{C}=90^{\circ}$, if $\mathrm{D}$ is the mid-point of $\mathrm{BC}$, prove that $\mathrm{AB}^{2}=4 \mathrm{AD}^{2}-3 \mathrm{AC}^{2}$. Solution: $\triangle \mathrm{ABC}$ is a right-angled triangle with $\angle \mathrm{C}=90^{\circ}$. D is the mid-point of $\mathrm{BC}$. We need to prove that $A B^{2}=4 A D^{2}-3 A C^{2}$. Join AD. Since D is the midpoint of the side BC, we get BD = DC $\therefore B C=2 D C$ Using Pythagoras...
Read More →Show that
Question: The value of $\int_{0}^{1} \tan ^{-1}\left(\frac{2 x-1}{1+x-x^{2}}\right) d x$ is A. 1 B. 0 C. $-1$ D. $\frac{\pi}{4}$ Solution: Let $I=\int_{0}^{1} \tan ^{-1}\left(\frac{2 x-1}{1+x-x^{2}}\right) d x$ $\Rightarrow I=\int_{0}^{1} \tan ^{-1}\left(\frac{x-(1-x)}{1+x(1-x)}\right) d x$ $\Rightarrow I=\int_{0}^{1}\left[\tan ^{-1} x-\tan ^{-1}(1-x)\right] d x$ ...(1) $\Rightarrow I=\int_{0}^{1}\left[\tan ^{-1}(1-x)-\tan ^{-1}(1-1+x)\right] d x$ $\Rightarrow I=\int_{0}^{1}\left[\tan ^{-1}(1-x)...
Read More →Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are
Question: Two chords AB and CD of lengths 5 cm and 11 cm respectively of a circle are parallel to each other and are opposite side of its centre. If the distance between AB and CD is 6 cm, find the radius of the circle. Solution: DrawOMAB and ONCD. Join OB and OD. BM = AB/2 = 5/2[Perpendicular from the centre bisects the chord] ND = CD/2 = 11/2 Let ON be x, so OM will be 6 - x. ΔMOB $\mathrm{OM}^{2}+\mathrm{MB}^{2}=\mathrm{OB}^{2}$ $(6-x)^{2}+(5 / 2)^{2}=O B^{2}$ $36+x^{2}-12 x+25 / 4=O B^{2} \l...
Read More →Calculate the height of an equilateral triangle each of whose sides measures 12 cm.
Question: Calculate the height of an equilateral triangle each of whose sides measures 12 cm. Solution: We are asked to find the height of the equilateral triangle. Let us draw the figure. Let us draw the altitude AD. We know that altitude is also median of the equilateral triangle. Therefore, $B D=D C=6 \mathrm{~cm}$ In right angled triangle ADB, we will Pythagoras theorem, as shown below, $A B^{2}=A D^{2}+B D^{2}$ Now we will substitute the values. $12^{2}=A D^{2}+6^{2}$ $144=A D^{2}+36$ $A D^...
Read More →Prove the following identities (1-16)
Question: Prove the following identities (1-16) $\frac{1-\sin x \cos x}{\cos x(\sec x-\operatorname{cosec} x)} \cdot \frac{\sin ^{2} x-\cos ^{2} x}{\sin ^{3} x+\cos ^{3} x}=\sin x$ Solution: LHS $=\frac{1-\sin x \cos x}{\cos x(\sec x-\operatorname{cosec} x)} \times \frac{\sin ^{2} x-\cos ^{2} x}{\sin ^{3} x+\cos ^{3} x}$ $=\frac{1-\sin x \cos x}{\cos x\left(\frac{1}{\cos x}-\frac{1}{\sin x}\right)} \times \frac{(\sin x)^{2}-(\cos x)^{2}}{(\sin x)^{3}+(\cos x)^{3}}$ $=\frac{1-\sin x \cos x}{\cos ...
Read More →Show that
Question: If $f(a+b-x)=f(x)$, then $\int_{a}^{b} x f(x) d x$ is equal to A. $\frac{a+b}{2} \int_{a}^{b} f(b-x) d x$ B. $\frac{a+b}{2} \int_{a}^{b} f(b+x) d x$ C. $\frac{b-a}{2} \int_{a}^{b} f(x) d x$ D. $\frac{a+b}{2} \int_{a}^{b} f(x) d x$ Solution: Let $I=\int_{0}^{b} x f(x) d x$ ...(1) $I=\int_{a}^{b}(a+b-x) f(a+b-x) d x$ $\left(\int_{a}^{b} f(x) d x=\int_{a}^{b} f(a+b-x) d x\right)$ $\Rightarrow I=\int_{a}^{b}(a+b-x) f(x) d x$ $\Rightarrow I=(a+b) \int_{a}^{b} f(x) d x \quad-I$ $[$ Using $(1...
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