Prove the following trigonometric identities.
Question: Prove the following trigonometric identities. $\frac{\tan A}{\left(1+\tan ^{2} A\right)^{2}}+\frac{\cot A}{\left(1+\cot ^{2} A\right)}=\sin A \cos A$ Solution: We have to prove $\frac{\tan A}{\left(1+\tan ^{2} A\right)^{2}}+\frac{\cot A}{\left(1+\cot ^{2} A\right)^{2}}=\sin A \cos A$ We know that, $\sin ^{2} A+\cos ^{2} A=1$. So, $\frac{\tan A}{\left(1+\tan ^{2} A\right)^{2}}+\frac{\cot A}{\left(1+\cot ^{2} A\right)^{2}}$ $=\frac{\tan A}{\left(\sec ^{2} A\right)^{2}}+\frac{\cot A}{\lef...
Read More →If 4 sin
Question: If $4 \sin ^{2} x=1$, then the values of $x$ are (a) $2 n \pi \pm \frac{\pi}{3}, n \in Z$ (b) $n \pi \pm \frac{\pi}{3}, n \in Z$ (c) $n \pi \pm \frac{\pi}{6}, n \in Z$ (d) $2 n \pi \pm \frac{\pi}{6}, n \in Z$ Solution: (c) $n \pi \pm \frac{\pi}{6}, n \in Z$ Given; $4 \sin ^{2} x=1$ $\Rightarrow \sin ^{2} x=\frac{1}{4}$ $\Rightarrow \sin x=\frac{1}{2} \quad$ or $\quad \sin x=-\frac{1}{2}$ $\Rightarrow \sin x=\sin \frac{\pi}{6} \quad$ or $\sin x=\sin \left(-\frac{\pi}{6}\right)$ $\Righta...
Read More →Prove the following trigonometric identities.
Question: Prove the following trigonometric identities. $\frac{\sin A}{\sec A+\tan A-1}+\frac{\cos A}{\operatorname{cosec} A+\cot A-1}=1$ Solution: We have to prove $\frac{\sin A}{\sec A+\tan A-1}+\frac{\cos A}{\operatorname{cosec} A+\cot A-1}=1$ We know that, $\sin ^{2} A+\cos ^{2} A=1$ So, $\frac{\sin A}{\sec A+\tan A-1}+\frac{\cos A}{\operatorname{cosec} A+\cot A-1}$ $=\frac{\sin A}{\frac{1}{\cos A}+\frac{\sin A}{\cos A}-1}+\frac{\cos A}{\frac{1}{\sin A}+\frac{\cos A}{\sin A}-1}$ $=\frac{\sin...
Read More →Simplify:
Question: Simplify: (i) $\frac{6^{1 / 4}}{6^{1 / 5}}$ (ii) $\frac{8^{1 / 2}}{8^{2 / 3}}$ (iii) $\frac{5^{6 / 7}}{5^{2 / 3}}$ Solution: $(\mathrm{i}) \frac{6^{\frac{1}{4}}}{6^{\frac{1}{5}}}=6^{\frac{1}{4}-\frac{1}{5}}=6^{\frac{5-4}{20}}=6^{\frac{1}{20}} \quad\left(\frac{a^{m}}{a^{n}}=a^{m-n}\right)$ $(\mathrm{ii}) \frac{8^{\frac{1}{2}}}{8^{\frac{2}{3}}}=8^{\frac{1}{2}-\frac{2}{3}}=8^{\frac{3-4}{6}}=8^{\frac{-1}{6}}$ $(\mathrm{iii}) \frac{5^{\frac{6}{7}}}{5^{\frac{2}{3}}}=5^{\frac{6}{7}-\frac{2}{3...
Read More →The smallest positive angle which satisfies the equation
Question: The smallest positive angle which satisfies the equation $2 \sin ^{2} x+\sqrt{3} \cos x+1=0$ is (a) $\frac{5 \pi}{6}$ (b) $\frac{2 \pi}{3}$ (c) $\frac{\pi}{3}$ (d) $\frac{\pi}{6}$ Solution: (a) $\frac{5 \pi}{6}$ Given; $2 \sin ^{2} x+\sqrt{3} \cos x+1=0$ $\Rightarrow 2\left(1-\cos ^{2} x\right)+\sqrt{3} \cos x+1=0$ $\Rightarrow 2-2 \cos ^{2} x+\sqrt{3} \cos x+1=0$ $\Rightarrow 2 \cos ^{2} x-\sqrt{3} \cos x-3=0$ $\Rightarrow 2 \cos ^{2} x-2 \sqrt{3} \cos x+\sqrt{3} \cos x-3=0$ $\Rightar...
Read More →Prove the following trigonometric identities.
Question: Prove the following trigonometric identities. (i) $\frac{\cos A \operatorname{cosec} A-\sin A \sec A}{\cos A+\sin A}=\operatorname{cosec} A-\sec A$ (ii) $\frac{\sin A}{\sec A+\tan A-1}+\frac{\cos A}{\operatorname{cosec} A+\cot A-1}=1$ Solution: (i) We have to prove $\frac{\cos A \operatorname{cosec} A-\sin A \sec A}{\cos A+\sin A}=\operatorname{cosec} A-\sec A$ So, $\frac{\cos A \operatorname{cosec} A-\sin A \sec A}{\cos A+\sin A}=\frac{\cos A \frac{1}{\sin A}-\sin A \frac{1}{\cos A}}{...
Read More →Simplify
Question: Simplify (i) $2^{\frac{2}{3}} \times 2^{\frac{1}{3}}$ (ii) $2^{\frac{2}{3}} \times 2^{\frac{1}{5}}$ (iii) $7^{\frac{5}{6}} \times 7^{\frac{2}{3}}$ (iv) $(1296)^{\frac{1}{4}} \times(1296)^{\frac{1}{2}}$ Solution: (i) $2^{\frac{2}{3}} \times 2^{\frac{1}{3}}$ $2^{\frac{2}{3}} \times 2^{\frac{1}{3}}=2^{\frac{2}{3}+\frac{1}{3}}$ $=2^{\frac{2+1}{3}}$ $=2^{\frac{3}{3}}$ $=2^{1}$ $=2$ (ii) $2^{\frac{2}{3}} \times 2^{\frac{1}{5}}$ $2^{\frac{2}{3}} \times 2^{\frac{1}{5}}=2^{\frac{2}{3}+\frac{1}{...
Read More →The general value of x satisfying the equation
Question: The general value of $x$ satisfying the equation $\sqrt{3} \sin x+\cos x=\sqrt{3}$ is given by (a) $x=n \pi+(-1)^{n} \frac{\pi}{4}+\frac{\pi}{3}, n \in Z$ (b) $x=n \pi+(-1)^{n} \frac{\pi}{3}+\frac{\pi}{6}, n \in Z$ (c) $x=n \pi \pm \frac{\pi}{6}, n \in Z$ (d) $x=n \pi \pm \frac{\pi}{3}, n \in Z$ Solution: (b) $x=n \pi+(-1)^{n} \frac{\pi}{3}+\frac{\pi}{6}, n \in Z$ Given: $\sqrt{3} \sin x+\cos x=\sqrt{3} \ldots$ (i) This equation is of the form $a \sin \theta+b \cos \theta=c$, where $a=...
Read More →The general value of x satisfying the equation
Question: The general value of $x$ satisfying the equation $\sqrt{3} \sin x+\cos x=\sqrt{3}$ is given by (a) $x=n \pi+(-1)^{n} \frac{\pi}{4}+\frac{\pi}{3}, n \in Z$ (b) $x=n \pi+(-1)^{n} \frac{\pi}{3}+\frac{\pi}{6}, n \in Z$ (c) $x=n \pi \pm \frac{\pi}{6}, n \in Z$ (d) $x=n \pi \pm \frac{\pi}{3}, n \in Z$ Solution: (b) $x=n \pi+(-1)^{n} \frac{\pi}{3}+\frac{\pi}{6}, n \in Z$ Given: $\sqrt{3} \sin x+\cos x=\sqrt{3} \ldots$ (i) This equation is of the form $a \sin \theta+b \cos \theta=c$, where $a=...
Read More →The number of solution in [0, π/2]
Question: The number of solution in $[0, \pi / 2]$ of the equation $\cos 3 x \tan 5 x=\sin 7 x$ is (a) 5 (b) 7 (c) 6 (d) none of these Solution: (c) 6 Given; $\cos 3 x \tan 5 x=\sin 7 x$ $\Rightarrow \cos (5 x-2 x) \tan 5 x=\sin (5 x+2 x)$ $\Rightarrow \tan 5 x=\frac{\sin (5 x+2 x)}{\cos (5 x-2 x)}$ $\Rightarrow \tan 5 x=\frac{\sin 5 x \cos 2 x+\cos 5 x \sin 2 x}{\cos 5 x \cos 2 x+\sin 5 x \sin 2 x}$ $\Rightarrow \frac{\sin 5 x}{\cos 5 x}=\frac{\sin 5 x \cos 2 x+\cos 5 x \sin 2 x}{\cos 5 x \cos ...
Read More →Prove the following trigonometric identities.
Question: Prove the following trigonometric identities. (secA cosecA) (1 + tanA+ cotA) = tanAsecA cotAcosecA Solution: We have to prove $(\sec A-\operatorname{cosec} A)(1+\tan A+\cot A)=\tan A \sec A-\cot A \operatorname{cosec} A$ We know that, $\sin ^{2} A+\cos ^{2} A=1$ So, $(\sec A-\operatorname{cosec} A)(1+\tan A+\cot A)=\left(\frac{1}{\cos A}-\frac{1}{\sin A}\right)\left(1+\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}\right)$ $=\left(\frac{\sin A-\cos A}{\sin A \cos A}\right)\left(\frac{\sin ...
Read More →The number of solution in [0, π/2]
Question: The number of solution in $[0, \pi / 2]$ of the equation $\cos 3 x \tan 5 x=\sin 7 x$ is (a) 5 (b) 7 (c) 6 (d) none of these Solution: (c) 6 Given; $\cos 3 x \tan 5 x=\sin 7 x$ $\Rightarrow \cos (5 x-2 x) \tan 5 x=\sin (5 x+2 x)$ $\Rightarrow \tan 5 x=\frac{\sin (5 x+2 x)}{\cos (5 x-2 x)}$ $\Rightarrow \tan 5 x=\frac{\sin 5 x \cos 2 x+\cos 5 x \sin 2 x}{\cos 5 x \cos 2 x+\sin 5 x \sin 2 x}$ $\Rightarrow \frac{\sin 5 x}{\cos 5 x}=\frac{\sin 5 x \cos 2 x+\cos 5 x \sin 2 x}{\cos 5 x \cos ...
Read More →Prove the following trigonometric identities.
Question: Prove the following trigonometric identities. $(\operatorname{cosec} \theta-\sec \theta)(\cot \theta-\tan \theta)=(\operatorname{cosec} \theta+\sec \theta)(\sec \theta \operatorname{cosec} \theta-2)$ Solution: We have to prove $(\operatorname{cosec} \theta-\sec \theta)(\cot \theta-\tan \theta)=(\operatorname{cosec} \theta+\sec \theta)(\sec \theta \operatorname{cosec} \theta-2)$ We know that, $\sin ^{2} \theta+\cos ^{2} \theta=1$ Consider the LHS. $(\operatorname{cosec} \theta-\sec \the...
Read More →A solution of the equation cos
Question: A solution of the equation $\cos ^{2} x+\sin x+1=0$, lies in the interval (a) $(-\pi / 4, \pi / 4)$ (b) $(\pi / 4,3 \pi / 4)$ (c) $(3 \pi / 4,5 \pi / 4)$ (d) $(5 \pi / 4,7 \pi / 4)$ Solution: (d) $(5 \pi / 4,7 \pi / 4)$ Given: $\cos ^{2} x+\sin x+1=0$ $\Rightarrow\left(1-\sin ^{2} x\right)+\sin x+1=0$ $\Rightarrow 1-\sin ^{2} x+\sin x+1=0$ $\Rightarrow \sin ^{2} x-\sin x-2=0$ $\Rightarrow \sin ^{2} x-2 \sin x+\sin x-2=0$ $\Rightarrow \sin x(\sin x-2)+1(\sin x-2)=0$ $\Rightarrow(\sin x-...
Read More →Prove the following trigonometric identities.
Question: Prove the following trigonometric identities. $(1+\cot A-\operatorname{cosec} A)(1+\tan A+\sec A)=2$ Solution: We have to prove $(1+\cot A-\operatorname{cosec} A)(1+\tan A+\sec A)=2$ We know that, $\sin ^{2} A+\cos ^{2} A=1$. So, $(1+\cot A-\operatorname{cosec} A)(1+\tan A+\sec A)=\left(1+\frac{\cos A}{\sin A}-\frac{1}{\sin A}\right)\left(1+\frac{\sin A}{\cos A}+\frac{1}{\cos A}\right)$ $=\left(\frac{\sin A+\cos A-1}{\sin A}\right)\left(\frac{\cos A+\sin A+1}{\cos A}\right)$ $=\frac{(\...
Read More →The general solution of the equation
Question: The general solution of the equation $7 \cos ^{2} x+3 \sin ^{2} x=4$ is (a) $x=2 n \pi \pm \frac{\pi}{6}, n \in Z$ (b) $x=2 n \pi \pm \frac{2 \pi}{3}, n \in Z$ (c) $x=n \pi \pm \frac{\pi}{3}, n \in Z$ (d) none of these Solution: (c) $x=n \pi \pm \frac{\pi}{3}, n \in Z$ Given: $7 \cos ^{2} x+3 \sin ^{2} x=4$ $\Rightarrow 7 \cos ^{2} x+3\left(1-\cos ^{2} x\right)=4$ $\Rightarrow 7 \cos ^{2} x+3-3 \cos ^{2} x=4$ $\Rightarrow 4 \cos ^{2} x+3=4$ $\Rightarrow 4\left(1-\cos ^{2} x\right)=3$ $...
Read More →Prove the following trigonometric identities.
Question: Prove the following trigonometric identities. (secA+ tanA 1) (secA tanA+ 1) = 2 tanA Solution: We have to prove $(\sec A+\tan A-1)(\sec A-\tan A+1)=2 \tan A$ We know that, $\sec ^{2} A-\tan ^{2} A=1$ So, we have $(\sec A+\tan A-1)(\sec A-\tan A+1)=\{\sec A+(\tan A-1)\}\{\sec A-(\tan A-1)\}$ $=\sec ^{2} A-(\tan A-1)^{2}$ $=\sec ^{2} A-\left(\tan ^{2} A-2 \tan A+1\right)$ $=\left(\sec ^{2} A-\tan ^{2} A\right)+2 \tan A-1$ So, we have $(\sec A+\tan A-1)(\sec A-\tan A+1)=1+2 \tan A-1$ $=2 ...
Read More →Simplify
Question: Simplify (i) $2^{\frac{2}{3}} \times 2^{\frac{1}{3}}$ (ii) $2^{\frac{2}{3}} \times 2^{\frac{1}{5}}$ (iii) $7^{\frac{5}{6}} \times 7^{\frac{2}{3}}$ (iv) $(1296)^{\frac{1}{4}} \times(1296)^{\frac{1}{2}}$ Solution: (i) $2^{\frac{2}{3}} \times 2^{\frac{1}{3}}$ $2^{\frac{2}{3}} \times 2^{\frac{1}{3}}=2^{\frac{2}{3}+\frac{1}{3}}$ $=2^{\frac{2+1}{3}}$ $=2^{\frac{3}{3}}$ $=2^{1}$ $=2$ (ii) $2^{\frac{2}{3}} \times 2^{\frac{1}{5}}$ $2^{\frac{2}{3}} \times 2^{\frac{1}{5}}=2^{\frac{2}{3}+\frac{1}{...
Read More →If a is any real number, the number of roots of cot
Question: If $a$ is any real number, the number of roots of $\cot x-\tan x=a$ in the first quadrant is (are). (a) 2 (b) 0 (c) 1 (d) none of these Solution: (c) 1 Given: $\cot x-\tan x=a$ $\Rightarrow \frac{1}{\tan x}-\tan x=a$ $\Rightarrow 1-\tan ^{2} x=a \tan x$ $\Rightarrow \tan ^{2} x+a \tan x-1=0$ It is a quadratic equation. If $\tan x=z$, then the equation becomes $z^{2}+a z-1=0$ $\Rightarrow z=\frac{-a \pm \sqrt{a^{2}+4}}{2}$ $\Rightarrow \tan x=\frac{-a \pm \sqrt{a^{2}+4}}{2}$ $\Rightarro...
Read More →If a is any real number, the number of roots of cot
Question: If $a$ is any real number, the number of roots of $\cot x-\tan x=a$ in the first quadrant is (are). (a) 2 (b) 0 (c) 1 (d) none of these Solution: (c) 1 Given: $\cot x-\tan x=a$ $\Rightarrow \frac{1}{\tan x}-\tan x=a$ $\Rightarrow 1-\tan ^{2} x=a \tan x$ $\Rightarrow \tan ^{2} x+a \tan x-1=0$ It is a quadratic equation. If $\tan x=z$, then the equation becomes $z^{2}+a z-1=0$ $\Rightarrow z=\frac{-a \pm \sqrt{a^{2}+4}}{2}$ $\Rightarrow \tan x=\frac{-a \pm \sqrt{a^{2}+4}}{2}$ $\Rightarro...
Read More →If tan px−tan qx=0,
Question: If $\tan p x-\tan q x=0$, then the values of $\theta$ form a series in (a) AP (b) GP (c) HP (d) none of these Solution: (a) AP Given: $\tan p x-\tan q x=0$ $\Rightarrow \tan p x=\tan q x$ $\Rightarrow \frac{\sin p x}{\cos p x}=\frac{\sin q x}{\cos q x}$ $\Rightarrow \sin p x \cos q x=\sin q x \cos p x$ $\Rightarrow \frac{1}{2}\left[\sin \left(\frac{p+q}{2}\right) x+\sin \left(\frac{p-q}{2}\right) x\right]=\frac{1}{2}\left[\sin \left(\frac{q+p}{2}\right) x+\sin \left(\frac{q-p}{2}\right...
Read More →If tan px−tan qx=0,
Question: If $\tan p x-\tan q x=0$, then the values of $\theta$ form a series in (a) AP (b) GP (c) HP (d) none of these Solution: (a) AP Given: $\tan p x-\tan q x=0$ $\Rightarrow \tan p x=\tan q x$ $\Rightarrow \frac{\sin p x}{\cos p x}=\frac{\sin q x}{\cos q x}$ $\Rightarrow \sin p x \cos q x=\sin q x \cos p x$ $\Rightarrow \frac{1}{2}\left[\sin \left(\frac{p+q}{2}\right) x+\sin \left(\frac{p-q}{2}\right) x\right]=\frac{1}{2}\left[\sin \left(\frac{q+p}{2}\right) x+\sin \left(\frac{q-p}{2}\right...
Read More →Solve this
Question: Evaluate $\frac{15}{\sqrt{10}+\sqrt{20}+\sqrt{40}-\sqrt{5}-\sqrt{80}}$, it being given that $\sqrt{5}=2.236$ and $\sqrt{10}=3.162$. Hint $\frac{15}{\sqrt{10}+\sqrt{20}+\sqrt{40}-\sqrt{5}-\sqrt{80}}=\frac{15}{\sqrt{10}+2 \sqrt{5}+2 \sqrt{10}-\sqrt{5}-4 \sqrt{5}}$ $=\frac{15}{3 \sqrt{10}-3 \sqrt{5}}=\frac{5}{\sqrt{10}-\sqrt{5}}$ Solution: $\frac{15}{\sqrt{10}+\sqrt{20}+\sqrt{40}-\sqrt{5}-\sqrt{80}}=\frac{15}{\sqrt{10}+2 \sqrt{5}+2 \sqrt{10}-\sqrt{5}-4 \sqrt{5}}$ $=\frac{15}{3 \sqrt{10}-3...
Read More →If cos x
Question: If $\cos x+\sqrt{3} \sin x=2$, then $x=$ (a) $\pi / 3$ (b) $2 \pi / 3$ (c) $4 \pi / 6$ (d) $5 \pi / 12$ Solution: (a) $\pi / 3$ Given: $\cos x+\sqrt{3} \sin x=2 \quad \ldots(\mathrm{i})$ This equation is of the form $a \cos x+b \sin x=c$, where $a=1, b=\sqrt{3}$ and $c=2$. Let: $a=r \cos \alpha$ and $b=\sin \alpha$ Now, $1=r \cos \alpha, \sqrt{3}=r \sin \alpha$ $\Rightarrow r=\sqrt{a^{2}+b^{2}}=\sqrt{1+3}=\sqrt{4}=2$ And, $\tan \alpha=\frac{b}{a}$ $\Rightarrow \tan \alpha=\frac{\sqrt{3...
Read More →Prove the following trigonometric identities.
Question: Prove the following trigonometric identities. (i) $\left(\frac{1+\sin \theta-\cos \theta}{1+\sin \theta+\cos \theta}\right)^{2}=\frac{1-\cos \theta}{1+\cos \theta}$ (ii) $\frac{1+\sec \theta-\tan \theta}{1+\sec \theta+\tan \theta}=\frac{1-\sin \theta}{\cos \theta}$ Solution: (i) In the given question, we need to prove $\left(\frac{1+\sin \theta-\cos \theta}{1+\sin \theta+\cos \theta}\right)^{2}=\frac{1-\cos \theta}{1+\cos \theta}$ Taking $\sin \theta$ common from the numerator and the ...
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