Simplify the following
Question: Simplify $\frac{7 \sqrt{3}}{\sqrt{10}+\sqrt{3}}-\frac{2 \sqrt{5}}{\sqrt{6}+\sqrt{5}}-\frac{3 \sqrt{2}}{\sqrt{15}+3 \sqrt{2}} .$ Solution: We have, $\frac{7 \sqrt{3}}{\sqrt{10}+\sqrt{3}}-\frac{2 \sqrt{5}}{\sqrt{6}+\sqrt{5}}-\frac{3 \sqrt{2}}{\sqrt{15}+3 \sqrt{2}}$$\ldots($ i) Now, $\frac{7 \sqrt{3}}{\sqrt{10}+\sqrt{3}} \times \frac{\sqrt{10}-\sqrt{3}}{\sqrt{10}-\sqrt{3}}=\frac{7 \sqrt{30}-21}{(\sqrt{10})^{2}-(\sqrt{3})^{2}}$ [by rationalisation] [using identity, $(a-b)(a+b)=a^{2}-b^{2}$...
Read More →Prove that:
Question: Prove that: $\left|\begin{array}{ccc}a^{2} b c a c+c^{2} \\ a^{2}+a b b^{2} a c \\ a b b^{2}+b c c^{2}\end{array}\right|=4 a^{2} b^{2} c^{2}$ Solution: Let LHS $=\Delta=\mid a^{2} \quad b c \quad a c+c^{2}$ $\begin{array}{lcc}a^{2}+a b b^{2} a c \\ a b b^{2}+b c c^{2}\end{array}$ $\Delta=a b c \mid a \quad c \quad a+c$ $\begin{array}{lcc}a+b b a \\ b b+c c \mid\end{array}$ [Taking out $a, b$ and $c$ common from $\mathrm{C}_{1}, \mathrm{C}_{2}$ and $\mathrm{C}_{3}$ ] $\begin{array}{llll...
Read More →If 3x = cosec θ and
Question: If $3 x=\operatorname{cosec} \theta$ and $\frac{3}{x}=\cot \theta$, than $3\left(x^{2}-\frac{1}{x^{2}}\right)=?$ (a) $\frac{1}{27}$ (b) $\frac{1}{81}$ (c) $\frac{1}{3}$ (d) $\frac{1}{9}$ Solution: (c) $\frac{1}{3}$ Given: $3 x=\operatorname{cosec} \theta$ and $\frac{3}{x}=\cot \theta$ Also, we can deduce that $x=\frac{\cos e c \theta}{3}$ and $\frac{1}{x}=\frac{\cot \theta}{3}$. So, substituting the values of $x$ and $\frac{1}{x}$ in the given expression, we get: $3\left(x^{2}-\frac{1}...
Read More →Prove the following
Question: Express $0.6+0 . \overline{7}+0.4 \overline{7}$ in the form $\frac{p}{q}$, where $p$ and $q$ are integers and $q \neq 0 .$ Solution: Let $x=0 . \overline{7}=0.777$ $\ldots(\mathrm{i})$ On multiplying both sides of Eq. (i) by 10, we get $10 x=7.77 \ldots$ ...(ii) On subtracting Eq. (i) from Eq. (ii), we get $10 x-x=(7.77 \ldots)-(0.77 \ldots)$ $\Rightarrow$ $9 x=7$ $\therefore$ $x=\frac{7}{9}$ Now, let $\quad y=0.47=0.4777 \ldots . \quad \ldots$ (iii) On multiplying both sides of Eq. (i...
Read More →If 2x = sec A and
Question: If $2 x=\sec A$ and $\frac{2}{x}=\tan A$, then $2\left(x^{2}-\frac{1}{x^{2}}\right)=?=?$ (a) $\frac{1}{2}$ (b) $\frac{1}{4}$ (c) $\frac{1}{8}$ (d) $\frac{1}{16}$ Solution: (a) $\frac{1}{2}$ Given: $2 x=\sec A$ and $\frac{2}{x}=\tan A$ Also, we can deduce that $\mathrm{x}=\frac{\sec A}{2}$ and $\frac{1}{x}=\frac{\tan A}{2}$. So, substituting the values of $x$ and $\frac{1}{x}$ in the given expression, we get: $2\left(x^{2}-\frac{1}{x^{2}}\right)=2\left(\left(\frac{\sec A}{2}\right)^{2}-...
Read More →Solve this
Question: If $3 \cot \theta=4$, then $\frac{(5 \sin \theta+3 \cos \theta)}{(5 \sin \theta-3 \cos \theta)}=?$ (a) $\frac{1}{3}$ (b) 3 (c) $\frac{1}{9}$ (d) 9 Solution: (d) 9 We have $\frac{(5 \sin \theta+3 \cos \theta)}{(5 \sin \theta-3 \cos \theta)}$. Dividing the numerator and denominator of the given expression by sin,we get: $\frac{\frac{1}{\sin \theta}(5 \sin \theta+3 \cos \theta)}{\frac{1}{\sin \theta}(5 \sin \theta-3 \cos \theta)}$ $=\frac{5+3 \cot \theta}{5-3 \cot \theta}$ $=\frac{5+4}{5-...
Read More →Simplify the following
Question: Simplify (i) $\left(1^{3}+2^{3}+3^{3}\right)^{\frac{1}{2}}$ (ii) $\left(\frac{3}{5}\right)^{4}\left(\frac{8}{5}\right)^{-12}\left(\frac{32}{5}\right)^{6}$ (iii) $\left(\frac{1}{27}\right)^{\frac{-2}{3}}$ (iv) $\left.\left[(625)^{-\frac{1}{2}}\right)^{-\frac{1}{4}}\right]^{2}$ (v) $\frac{9^{\frac{1}{3}} \times 27^{-\frac{1}{2}}}{3^{\frac{1}{6}} \times 3^{-\frac{2}{3}}}$ (vi) $64^{-\frac{1}{3}}\left[64^{\frac{1}{3}}-64^{\frac{2}{3}}\right]$ (vii) $\frac{8^{\frac{1}{3}} \times 16^{\frac{1...
Read More →Evaluate:
Question: Evaluate: (i) $\sqrt[3]{4^{3} \times 6^{3}}$ (ii) $\sqrt[3]{8 \times 17 \times 17 \times 17}$ (iii) $\sqrt[3]{700 \times 2 \times 49 \times 5}$ (iv) $125 \sqrt[3]{\alpha^{6}}-\sqrt[3]{125 \alpha^{6}}$ Solution: Property: For any two integers $a$ and $b, \sqrt[3]{a b}=\sqrt[3]{a} \times \sqrt[3]{b}$ (i) From the above property, we have: $\sqrt[3]{4^{3} \times 6^{3}}=\sqrt[3]{4^{3}} \times \sqrt[3]{6^{3}}=4 \times 6=24$ (ii) Use above property and proceed as follows: $\sqrt[3]{8 \times 1...
Read More →If cos A + cos2 A = 1, then (sin2 A + sin4 A) = ?
Question: If cosA+ cos2A= 1, then (sin2A+ sin4A) = ?(a) 1(b) 2(c) 3(d) 4 Solution: (a) 1 $\cos A+\cos ^{2} A=1$ $=\cos A=1-\cos ^{2} A$ $=\cos A=\sin ^{2} A \quad\left(\because 1-\cos ^{2} A=\sin ^{2}\right)$ $=\cos ^{2} A=\sin ^{4} A \quad$ (Squaring both sides) $=1-\sin ^{2} A=\sin ^{4} A$ $=\sin ^{4} A+\sin ^{2} A=1$...
Read More →Evaluate:
Question: Evaluate: (i) $\sqrt[3]{4^{3} \times 6^{3}}$ (ii) $\sqrt[3]{8 \times 17 \times 17 \times 17}$ (iii) $\sqrt[3]{700 \times 2 \times 49 \times 5}$ (iv) $125 \sqrt[3]{\alpha^{6}}-\sqrt[3]{125 \alpha^{6}}$ Solution: Property: For any two integers $a$ and $b, \sqrt[3]{a b}=\sqrt[3]{a} \times \sqrt[3]{b}$ (i) From the above property, we have: $\sqrt[3]{4^{3} \times 6^{3}}=\sqrt[3]{4^{3}} \times \sqrt[3]{6^{3}}=4 \times 6=24$ (ii) Use above property and proceed as follows: $\sqrt[3]{8 \times 1...
Read More →If in ΔABC, ∠C = 90°. Then, the value of cos(A + B) is
Question: If in ΔABC, C= 90. Then, the value of cos(A + B) is(a) 0(b) 1 (c) $\frac{1}{2}$ (d) $\frac{3}{5}$ Solution: In ∆ABC,A + B + C = 180 (Angle sum property of triangle)⇒ A + B + 90 = 180 (C = 90)⇒ A + B = 180 90 = 90 cos(A + B) = cos90 = 0Thus, the value ofcos(A + B) is 0.Hence, the correct answer is option (a)....
Read More →Solve this
Question: If $5 \tan \theta=3$ then $\left(\frac{5 \sin \theta-\cos \theta}{5 \sin \theta+\cos \theta}\right)=?$ (a) $\frac{2}{3}$ (b) $\frac{1}{3}$ (c) $\frac{1}{2}$ (d) $\frac{3}{5}$ Solution: $5 \tan \theta=3$ $\Rightarrow 5 \times \frac{\sin \theta}{\cos \theta}=3$ $\Rightarrow 5 \sin \theta=3 \cos \theta \quad \ldots \ldots(1)$ $\therefore \frac{5 \sin \theta-\cos \theta}{5 \sin \theta+\cos \theta}$ $=\frac{3 \cos \theta-\cos \theta}{3 \cos \theta+\cos \theta} \quad[$ Using $(1)]$ $=\frac{2...
Read More →Prove that:
Question: Prove that: $\left|\begin{array}{ccc}1 a^{2}+b c a^{3} \\ 1 b^{2}+c a b^{3} \\ 1 c^{2}+a b c^{3}\end{array}\right|=-(a-b)(b-c)(c-a)\left(a^{2}+b^{2}+c^{2}\right)$ Solution: Let LHS $=\Delta=\mid \begin{array}{ll}1 \mathrm{a}^{2}+\mathrm{bc} \mathrm{a}^{3}\end{array}$ $1 \quad b^{2}+c a \quad b^{3}$ $\begin{array}{lcc}1 c^{2}+a b c^{3}\end{array}$ $\Rightarrow \Delta=\mid 0 \quad\left(a^{2}+b c\right)-\left(b^{2}+c a\right) \quad a^{3}-b^{3}$ $0 \quad\left(b^{2}+c a\right)-\left(c^{2}+a...
Read More →Find the cube root of each of the following numbers:
Question: Find the cube root of each of the following numbers: (i) 8 125 (ii) 1728 216 (iii) 27 2744 (iv) 729 15625 Solution: Property:For any two integersaandb, $\sqrt[3]{a b}=\sqrt[3]{a} \times \sqrt[3]{b}$ (i)From the above property, we have: $\sqrt[3]{8 \times 125}=\sqrt[3]{8} \times \sqrt[3]{125}=\sqrt[3]{2 \times 2 \times 2} \times \sqrt[3]{5 \times 5 \times 5}=2 \times 5=10$ (ii)From the above property, we have: $\sqrt[3]{-1728 \times 216}$ $=\sqrt[3]{-1728} \times \sqrt[3]{216}$ $=-\sqr...
Read More →Rationalise the denominator in each of the following
Question: Rationalise the denominator in each of the following and hence evaluate by taking $\sqrt{2}=1.414, \sqrt{3}=1.732$ and $\sqrt{5}=2.236$ upto three places of decimal. (i) $\frac{4}{\sqrt{3}}$ (ii) $\frac{6}{\sqrt{6}}$ (iii) $\frac{\sqrt{10}-\sqrt{5}}{2}$ (iv) $\frac{\sqrt{2}}{2+\sqrt{2}}$ (v) $\frac{1}{\sqrt{3}+\sqrt{2}}$ Solution: (i) Let $E=\frac{4}{\sqrt{3}}$ For rationalising the denominator, multiplying numerator and denominator by $\sqrt{3}$, we get $E=\frac{4}{\sqrt{3}} \times \f...
Read More →Solve this
Question: If $\tan \theta=\frac{8}{15}$ then $\operatorname{cosec} \theta=?$ (a) $\frac{17}{8}$ (b) $\frac{8}{17}$ (c) $\frac{15}{17}$ (d) $\frac{17}{15}$ Solution: $\tan \theta=\frac{8}{15}$ $\Rightarrow \frac{1}{\tan \theta}=\frac{15}{8}$ $\Rightarrow \cot \theta=\frac{15}{8} \quad \ldots \ldots(1)$ Now, $\operatorname{cosec}^{2} \theta=1+\cot ^{2} \theta$ $\Rightarrow \operatorname{cosec}^{2} \theta=1+\left(\frac{15}{8}\right)^{2} \quad$ [Using (1)] $\Rightarrow \operatorname{cosec}^{2} \thet...
Read More →Prove the following
Question: If $a=2+\sqrt{3}$, then find the value of $\left(a-\frac{1}{a}\right)$. Solution: We have, $a=2+\sqrt{3}$ Then, $\frac{1}{a}=\frac{1}{2+\sqrt{3}} \times \frac{2-\sqrt{3}}{2-\sqrt{3}}$ [multiplying numerator and denominator by $(2-\sqrt{3})$ ] $=\frac{2-\sqrt{3}}{(2)^{2}-(\sqrt{3})^{2}}=\frac{2-\sqrt{3}}{4-3}=2-\sqrt{3}$ [using identity, $(a+b)(a-b)=a^{2}-b^{2}$ ] $\Rightarrow \quad \frac{1}{a}=2-\sqrt{3}$ $\therefore \quad a-\frac{1}{a}=(2+\sqrt{3})-(2-\sqrt{3})=2+\sqrt{3}-2+\sqrt{3}=2...
Read More →Show that:
Question: Show that: (i) $\sqrt[3]{27} \times \sqrt[3]{64}=\sqrt[3]{27 \times 64}$ (ii) $\sqrt[3]{64 \times 729}=\sqrt[3]{64} \times \sqrt[3]{729}$ (iv) $\sqrt[3]{-125 \times 216}=\sqrt[3]{-125} \times \sqrt[3]{216}$ (v) $\sqrt[3]{-125-1000}=\sqrt[3]{-125} \times \sqrt[3]{-1000}$ Solution: (i) LHS $=\sqrt[3]{27} \times \sqrt[3]{64}=\sqrt[3]{3 \times 3 \times 3} \times \sqrt[3]{4 \times 4 \times 4}=3 \times 4=12$ RHS $=\sqrt[3]{27 \times 64}=\sqrt[3]{3 \times 3 \times 3 \times 4 \times 4 \times 4...
Read More →Show that:
Question: Show that: (i) $\sqrt[3]{27} \times \sqrt[3]{64}=\sqrt[3]{27 \times 64}$ (ii) $\sqrt[3]{64 \times 729}=\sqrt[3]{64} \times \sqrt[3]{729}$ (iv) $\sqrt[3]{-125 \times 216}=\sqrt[3]{-125} \times \sqrt[3]{216}$ (v) $\sqrt[3]{-125-1000}=\sqrt[3]{-125} \times \sqrt[3]{-1000}$ Solution: (i) LHS $=\sqrt[3]{27} \times \sqrt[3]{64}=\sqrt[3]{3 \times 3 \times 3} \times \sqrt[3]{4 \times 4 \times 4}=3 \times 4=12$ RHS $=\sqrt[3]{27 \times 64}=\sqrt[3]{3 \times 3 \times 3 \times 4 \times 4 \times 4...
Read More →Solve this
Question: If $\tan \theta=\frac{1}{\sqrt{7}}$, then $\frac{\left(\operatorname{cosec}^{2} \theta-\sec ^{2} \theta\right)}{\left(\operatorname{cosec}^{2} \theta+\sec ^{2} \theta\right)}=?=?$ (a) $\frac{-2}{3}$ (b) $\frac{-3}{4}$ (c) $\frac{2}{3}$ (d) $\frac{3}{4}$ Solution: (d) $\frac{3}{4}$ $=\frac{\operatorname{cosec}^{2} \theta-\sec ^{2} \theta}{\operatorname{cosec}^{2} \theta+\sec ^{2} \theta}$ $=\frac{\sin ^{2} \theta\left(\frac{1}{\sin ^{2} \theta}-\frac{1}{\cos ^{2} \theta}\right)}{\sin ^{...
Read More →Find the values of a and b in each of the following
Question: Find the values of $a$ and $b$ in each of the following (i) $\frac{5+2 \sqrt{3}}{7+4 \sqrt{3}}=a-6 \sqrt{3}$ (ii) $\frac{3-\sqrt{5}}{3+2 \sqrt{5}}=a \sqrt{5}-\frac{19}{11}$ (iii) $\frac{\sqrt{2}+\sqrt{3}}{3 \sqrt{2}-2 \sqrt{3}}=2-b \sqrt{6}$ (iv) $\frac{7+\sqrt{5}}{7-\sqrt{5}}-\frac{7-\sqrt{5}}{7+\sqrt{5}}=a+\frac{7}{11} \sqrt{5} b$ Solution: (i) We have, $\frac{5+2 \sqrt{3}}{7+4 \sqrt{3}}=a-6 \sqrt{3}$ For rationalising the above equation, we multiply numerator and denominator of LHS ...
Read More →Solve this
Question: If $\tan \theta=\frac{a}{b}$, then $\frac{(\cos \theta+\sin \theta)}{(\cos \theta-\sin \theta)}=?$ (a) $\frac{a+b}{a-b}$ (b) $\frac{a-b}{a+b}$ (c) $\frac{b+a}{b-a}$ (d) $\frac{b-a}{b+a}$ Solution: (c) $\frac{b+a}{b-a}$ Given: $\tan \theta=\frac{a}{b}$ Now, $\frac{(\cos \theta+\sin \theta)}{(\cos \theta-\sin \theta)}$ $=\frac{(1+\tan \theta)}{(1-\tan \theta)} \quad[$ Dividing the numerator and denominator by $\cos \theta]$ $=\frac{\left(1+\frac{a}{b}\right)}{\left(1-\frac{a}{b}\right)}$...
Read More →Find the cube roots of each of the following integers:
Question: Find the cube roots of each of the following integers: (i) 125 (ii) 5832 (iii) 2744000 (iv) 753571 (v) 32768 Solution: (i)We have: $\sqrt[3]{-125}=-\sqrt[3]{125}=-\sqrt[3]{5 \times 5 \times 5}=-5$ (ii)We have: $\sqrt[3]{-5832}=-\sqrt[3]{5832}$ To find the cube root of 5832, we use the method of unit digits. Let us consider the number 5832. The unit digit is 2; therefore the unit digit in the cube root of 5832 will be 8. After striking out the units, tens and hundreds digits of the give...
Read More →If (tan θ + cot θ) = 5, then (tan2 θ + cot2 θ) = ?
Question: If (tan + cot ) = 5, then (tan2 + cot2) = ?(a) 23(b) 24(c) 25(d) 27 Solution: (d) 23We have (tan+cot) = 5Squaring both sides, we get:(tan+cot )2= 52⇒ tan2+ cot2+ 2 tancot= 25 $\Rightarrow \tan ^{2} \theta+\cot ^{2} \theta+2=25 \quad\left[\because \tan \theta=\frac{1}{\cot \theta}\right]$ ⇒ tan2 + cot2 = 25 2 = 23...
Read More →Prove that:
Question: Prove that: $\left|\begin{array}{lll}a^{2} a^{2}-(b-c)^{2} b c \\ b^{2} b^{2}-(c-a)^{2} c a \\ c^{2} c^{2}-(a-b)^{2} a b\end{array}\right|=(a-b)(b-c)(c-a)(a+b+c)\left(a^{2}+b^{2}+c^{2}\right)$ Solution: Let LHS $=\Delta=\mid \mathrm{a}^{2} \quad \mathrm{a}^{2}-(\mathrm{b}-\mathrm{c})^{2} \quad \mathrm{bc}$ $b^{2} \quad b^{2}-(c-a)^{2} \quad c a$ $c^{2} \quad c^{2}-(a-b)^{2} \quad a b \mid$ $\Rightarrow \Delta=\mid \mathrm{a}^{2}-(\mathrm{b}-\mathrm{c})^{2} \quad \mathrm{bc}$ $\begin{ar...
Read More →