If $u=\cot ^{-1} \sqrt{\tan \theta}-\tan ^{-1} \sqrt{\tan \theta}$ then, $\tan \left(\frac{\pi}{4}-\frac{u}{2}\right)=$
[question] Question. If $u=\cot ^{-1} \sqrt{\tan \theta}-\tan ^{-1} \sqrt{\tan \theta}$ then, $\tan \left(\frac{\pi}{4}-\frac{u}{2}\right)=$ (a) $\sqrt{\tan \theta}$ (b) $\sqrt{\cot \theta}$ (c) $\tan \theta$ (d) $\cot \theta$ [/question] [solution] Solution: (a) $\sqrt{\tan \theta}$ Let $y=\sqrt{\tan \theta}$ Then, $u=\cot ^{-1} \sqrt{\tan \theta}-\tan ^{-1} \sqrt{\tan \theta}$ $\Rightarrow u=\cot ^{-1} y-\tan ^{-1} y$ $\Rightarrow u=\frac{\pi}{2}-2 \tan ^{-1} y \quad\left[\because \tan ^{-1} x...
Read More →In the given figure, PQ > PR and QS and RS are the bisectors of ∠Q and
Question: In the given figure,PQPRandQSandRSare the bisectors ofQand Rrespectively. Show thatSQSR. Solution: Since the angle opposite to the longer side is greater, we have: $P QP R$ $\Rightarrow \angle R\angle Q$ $\Rightarrow \frac{1}{2} \angle R\frac{1}{2} \angle Q$ $\Rightarrow \angle S R Q\angle R Q S$ $\Rightarrow Q SS R$ $\therefore S QS R$...
Read More →In ΔABC, ∠B = 35°, ∠C = 65° and the bisector of ∠BAC meets BC in X.
Question: In ΔABC, B= 35, C= 65 and the bisector of BACmeetsBCinX. ArrangeAX,BXandCXin descending order. Solution: Given:In ΔABC, B= 35, C= 65 and the bisector of BACmeetsBCinX. In $\Delta A B X$, $\because \angle B A X\angle A B X$ $\therefore B XA X \quad \ldots(\mathrm{i})$ Similarly, in $\Delta A C X$, $\because \angle A C X\angle X A C$ $\therefore A XC X \quad \ldots$ (ii) From (i) and $(i i)$, we get $B XA XC X$...
Read More →The value of
Question: The value of $\sin ^{-1}\left(\cos \frac{33 \pi}{5}\right)$ is________________. Solution: $\sin ^{-1}\left(\cos \frac{33 \pi}{5}\right)$ $=\sin ^{-1}\left[\cos \left(6 \pi+\frac{3 \pi}{5}\right)\right]$ $=\sin ^{-1}\left(\cos \frac{3 \pi}{5}\right)$ $=\sin \left[\sin \left(\frac{\pi}{2}-\frac{3 \pi}{5}\right)\right]$ $=\sin \left[\sin \left(-\frac{\pi}{10}\right)\right]$ $=-\frac{\pi}{10}$ Thus, the value of $\sin ^{-1}\left(\cos \frac{33 \pi}{5}\right)$ is $-\frac{\pi}{10}$. The value...
Read More →if the
Question: If $\tan ^{-1}(\cot \theta)=2 \theta$, then $\theta=$_________________. Solution: $\tan ^{-1}(\cot \theta)=2 \theta$ $\Rightarrow \tan ^{-1}\left[\tan \left(\frac{\pi}{2}-\theta\right)\right]=2 \theta$ $\Rightarrow \frac{\pi}{2}-\theta=2 \theta$ $\Rightarrow 3 \theta=\frac{\pi}{2}$ $\Rightarrow \theta=\frac{\pi}{6}$ Thus, the value of $\theta$ is $\frac{\pi}{6}$. If $\tan ^{-1}(\cot \theta)=2 \theta$, then $\theta=\frac{\pi}{6}$...
Read More →Solve the following
Question: (i) Which term of the A.P. 3, 8, 13, ... is 248? (ii) Which term of the A.P. 84, 80, 76, ... is 0? (iii) Which term of the A.P. 4, 9, 14, ... is 254? Solution: (i) 3, 8, 13... Here, we have: a = 3 $d=(8-3)=5$ Let $a_{n}=248$ $\Rightarrow a+(n-1) d=248$ $\Rightarrow 3+(n-1) 5=248$ $\Rightarrow(n-1) 5=245$ $\Rightarrow n-1=49$ $\Rightarrow n=50$ Hence, 248 is the 50th term of the given A.P. (ii) 84, 80, 76... Here, we have: a = 84 $d=(80-84)=-4$ Let $a_{n}=0$ $\Rightarrow a+(n-1) d=0$ $\...
Read More →In a quadrilateral ABCD, show that
Question: In a quadrilateral $\mathrm{ABCD}_{1}$ show that $(A B+B C+C D+D A)2(B D+A C)$ Solution: Given: QuadrilateralABCDTo prove:(AB + BC + CD + DA) 2(BD + AC).Proof: In ∆AOB, $O A+O BA B \quad \ldots \ldots(i)$ In ∆BOC, $O B+O CB C \quad \ldots \ldots(i i)$ In ∆COD, $O C+O DC D \quad \ldots \ldots($ iii $)$ In ∆AOD, $O D+O AA D \quad \ldots \ldots(i v)$ Adding $(i),(i i),(i i i)$ and $(i v)$, we get $2(O A+O B+O C+O D)(A B+B C+C D+D A)$ $\Rightarrow 2(O B+O D+O A+O C)(A B+B C+C D+D A)$ $\Rig...
Read More →The value of
Question: The value of $\tan ^{2}\left(\sec ^{-1} 3\right)+\cot ^{2}\left(\operatorname{cosec}^{-1} 4\right)$ is_________________. Solution: $\tan ^{2}\left(\sec ^{-1} 3\right)+\cot ^{2}\left(\operatorname{cosec}^{-1} 4\right)$ $=\sec ^{2}\left(\sec ^{-1} 3\right)-1+\operatorname{cosec}^{2}\left(\operatorname{cosec}^{-1} 4\right)-1$ $\left(1+\tan ^{2} \theta=\sec ^{2} \theta\right.$ and $\left.1+\cot ^{2} \theta=\operatorname{cosec}^{2} \theta\right)$ $=\left[\sec \left(\sec ^{-1} 3\right)\right...
Read More →If $x<0, y<0$ such that $x y=1$, then $\tan ^{-1} x+\tan ^{-1} y$ equals
[question] Question. If $x<0, y<0$ such that $x y=1$, then $\tan ^{-1} x+\tan ^{-1} y$ equals (a) $\frac{\pi}{2}$ (b) $-\frac{\pi}{2}$ (c) $-\pi$ (d) none of these [/question] [solution] Solution: (b) $-\frac{\pi}{2}$ We know that $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$ $x<0, y<0$ such that $x y=1$ Let $x=-a$ and $y=-b$, where $a$ and $b$ both are positive. $\therefore \tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$ $=\tan ^{-1}\left(\frac{-a-a}{1...
Read More →Find the indicated terms in each of the following sequences whose nth terms are:
Question: Find the indicated terms in each of the following sequences whose nth terms are: (a) $a_{n}=5 n-4 ; a_{12}$ and $a_{15}$ (b) $a_{n}=\frac{3 n-2}{4 n+5} ; a_{7}$ and $a_{8}$ (c) $a_{n}=n(n-1)(n-2) ; a_{5}$ and $a_{8}$ (d) $a_{n}=(n-1)(2-n)(3+n) ; a_{1}, a_{2}, a_{3}$ (e) $a_{n}=(-1)^{n} n ; a_{3}, a_{5}, a_{8}$ Solution: Here, we are given thenthterm for various sequences. We need to find the indicated terms of the A.P. (i) $a_{n}=5 n-4$ We need to findand Now, to findterm we use, we ge...
Read More →Find the indicated terms in each of the following sequences whose nth terms are:
Question: Find the indicated terms in each of the following sequences whose nth terms are: (a) $a_{n}=5 n-4 ; a_{12}$ and $a_{15}$ (b) $a_{n}=\frac{3 n-2}{4 n+5} ; a_{7}$ and $a_{8}$ (c) $a_{n}=n(n-1)(n-2) ; a_{5}$ and $a_{8}$ (d) $a_{n}=(n-1)(2-n)(3+n) ; a_{1}, a_{2}, a_{3}$ (e) $a_{n}=(-1)^{n} n ; a_{3}, a_{5}, a_{8}$ Solution: Here, we are given thenthterm for various sequences. We need to find the indicated terms of the A.P. (i) $a_{n}=5 n-4$ We need to findand Now, to findterm we use, we ge...
Read More →Solve the following
Question: If the sequence an is an A.P., show that am+n+amn= 2am. Solution: Let the sequence an be an A.P. with the first term beingAand the common difference beingD. To prove:am+n+amn= 2am LHS:am+n+amn $=A+(m+n-1) D+A+(m-n-1) D \quad\left\{\because a_{n}=a+(n-1) d\right\}$ $=A+m D+n D-D+A+m D-n D-D$ $=2 A+2 m D-2 D \quad \ldots(\mathrm{i})$ RHS: 2am $=2[A+(m-1) D]$ $=2 A+2 m D-2 D \quad \ldots(\mathrm{ii})$ From (i) and (ii), we get: LHS = RHS Hence, proved....
Read More →The value of
Question: The value of $\tan \left[\cos ^{-1}\left\{\sin \left(\cot ^{-1} 1\right)\right\}\right]$ is___________________ Solution: $\tan \left\{\cos ^{-1}\left[\sin \left(\cot ^{-1} 1\right)\right]\right\}$ $=\tan \left\{\cos ^{-1}\left[\sin \left(\frac{\pi}{4}\right)\right]\right\}$ $\left(\cot \frac{\pi}{4}=1 \Rightarrow \cot ^{-1} 1=\frac{\pi}{4}\right)$ $=\tan \left\{\cos ^{-1}\left(\frac{1}{\sqrt{2}}\right)\right\}$ $=\tan \frac{\pi}{4}$ $\left(\cos \frac{\pi}{4}=\frac{1}{\sqrt{2}} \Rightar...
Read More →Find:
Question: Find: (i) 10th term of the A.P. 1, 4, 7, 10, ... (ii) 18th term of the A.P. $\sqrt{2}, 3 \sqrt{2}, 5 \sqrt{2}, \ldots$ (iii) nth term of the A.P. $13,8,3,-2, \ldots$ Solution: (i) $1,4,7,10 \ldots$ We have; $a=1$ $d=4-1=3$ $a_{10}=a+(10-1) d \quad\left[a_{n}=a+(n-1) d\right]$ $=a+9 d$ $=1+9 \times 3$ $=28$ (ii) $\sqrt{2}, 3 \sqrt{2}, 5 \sqrt{2} \ldots$ We have; $a=\sqrt{2}$ $d=3 \sqrt{2}-\sqrt{2}=2 \sqrt{2}$ $a_{18}=a+(18-1) d \quad\left[a_{n}=a+(n-1) d\right]$ $=a+17 d$ $=\sqrt{2}+17(...
Read More →In a quadrilateral ABCD, show that
Question: In a quadrilateralABCD, show that (AB + BC + CD + DA) (AC + BD). Solution: Given: QuadrilateralABCDTo prove:(AB + BC + CD + DA) (AC + BD)Proof: In $\Delta A B C$, $A B+B CA C \quad \ldots$ (i) In $\triangle C A D$, $C D+A DA C \quad \ldots$ (ii) In $\Delta B A D$, $A B+A DB D \quad \ldots$ (iii) In $\Delta B C D$ $B C+C DB D \quad \ldots$ (iv) Adding (i), (ii), (iii) and (iv), we get2(AB+BC+CD+DA) 2(AC+BD)Hence, (AB+BC+CD+DA) (AC+BD)....
Read More →The value of
Question: The value of $\cos ^{-1}\left\{\sin \left(\cos ^{-1} \frac{1}{2}\right)\right\}$ is________________________. Solution: $\cos ^{-1}\left[\sin \left(\cos ^{-1} \frac{1}{2}\right)\right]$ $=\cos ^{-1}\left[\sin \left(\frac{\pi}{3}\right)\right]$ $\left(\cos \frac{\pi}{3}=\frac{1}{2} \Rightarrow \cos ^{-1} \frac{1}{2}=\frac{\pi}{3}\right)$ $=\cos ^{-1}\left(\frac{\sqrt{3}}{2}\right)$ $=\frac{\pi}{6}$ $\left(\cos \frac{\pi}{6}=\frac{\sqrt{3}}{2} \Rightarrow \cos ^{-1} \frac{\sqrt{3}}{2}=\fr...
Read More →AB and CD are respectively the smallest and largest sides of a quadrilateral ABCD
Question: ABandCDare respectively the smallest and largest sides of a quadrilateralABCD. Show that A CandB D. Solution: Given: InquadrilateralABCD,ABandCDare respectively the smallest and largest sides.To prove:(i) A C(ii)B D Construction: JoinAC.Proof: In $\Delta A B C$, $\because B CA B \quad$ (Given, $A B$ is the smallest side) $\therefore \angle 1\angle 2 \quad \ldots(1)$ In $\Delta A D C$, $\because C DA D \quad$ (Given, $C D$ is the largest side) $\therefore \angle 3\angle 4 \quad \ldots(...
Read More →if
Question: If $\sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z=-\frac{3 \pi}{2}$, then $x y z=$__________________. Solution: We know $-\frac{\pi}{2} \leq \sin ^{-1} a \leq \frac{\pi}{2}$, for all $a \in[-1,1]$ So, the minimum value of $\sin ^{-1} a$ is $-\frac{\pi}{2}$. Now, $\sin ^{-1} x+\sin ^{-1} y+\sin ^{-1} z=-\frac{3 \pi}{2} \quad$ (Given) This is possible if $\sin ^{-1} x=-\frac{\pi}{2}, \sin ^{-1} y=-\frac{\pi}{2}$ and $\sin ^{-1} z=-\frac{\pi}{2}$ $\Rightarrow x=-1, y=-1$ and $z=-1$ $\therefore x...
Read More →If 18th and 11th term of an A.P. are in the ratio 3 : 2, then its 21st and 5th terms are in the ratio
Question: If $18^{\text {th }}$ and $11^{\text {th }}$ term of an A.P. are in the ratio $3: 2$, then its $21^{\text {st }}$ and $5^{\text {th }}$ terms are in the ratio (a) 3 : 2(b) 3 : 1(c) 1 : 3(d) 2 : 3 Solution: In the given problem, we are given an A.P whose 18thand 11thterm are in the ratio 3:2 We need to find the ratio of its 21stand 5thterms Now, using the formula $a_{n}=a+(n-1) d$ Where, $a=$ first tem of the A.P $n=$ number of terms $d=$ common difference of the A.P So, $a_{18}=a+(18-1...
Read More →if
Question: If $\tan ^{-1} x-\cot ^{-1} x=\tan ^{-1} \sqrt{3}$, then $x=$_______________________. Solution: $\tan ^{-1} x-\cot ^{-1} x=\tan ^{-1} \sqrt{3}$ $\Rightarrow \tan ^{-1} x-\left(\frac{\pi}{2}-\tan ^{-1} x\right)=\frac{\pi}{3}$ $\left(\tan ^{-1} x+\cot ^{-1} x=\frac{\pi}{2}\right)$ $\Rightarrow 2 \tan ^{-1} x=\frac{\pi}{2}+\frac{\pi}{3}$ $\Rightarrow \tan ^{-1} x=\frac{\pi}{4}+\frac{\pi}{6}$ $\Rightarrow x=\tan \left(\frac{\pi}{4}+\frac{\pi}{6}\right)$ $\Rightarrow x=\frac{\tan \frac{\pi}...
Read More →If the nth term of an A.P. is 2n + 1, then the sum of first n terms of the A.P. is
Question: If the nth term of an A.P. is 2n+ 1, then the sum of firstnterms of the A.P. is (a)n(n 2)(b)n(n+ 2)(c)n(n+ 1)(d)n(n 1) Solution: Here, we are given an A.P. whose $n^{\text {th }}$ term is given by the following expression, $a_{n}=2 n+1$. We need to find the sum of first $n$ terms. So, here we can find the sum of the $n$ terms of the given A.P., using the formula, $S_{n}=\left(\frac{n}{2}\right)(a+l)$ Where,a= the first term l= the last term So, for the given A.P, The first term (a) wil...
Read More →In the given figure, ∠B < ∠A and ∠C < ∠D. Show that AD > BC.
Question: In the given figure, B Aand C D. Show thatADBC. Solution: Given:B Aand C DTo prove:ADBCProof: In $\Delta A O B$, $\angle B\angle A$ $\Rightarrow A O In $\Delta C O D$, $\angle C\angle D$ $\Rightarrow O D Adding (1) and (2), we get $A O+O D $\therefore A D...
Read More →The sum of n terms of an A.P. is 3n2 + 5n, then 164 is its
Question: The sum of $n$ terms of an A.P. is $3 n^{2}+5 n$, then 164 is its (a) 24thterm(b) 27thterm(c) 26thterm(d) 25thterm Solution: Here, the sum of firstnterms is given by the expression, We need to find which term of the A.P. is 164. Let us take 164 as thenthterm. So we know that thenthterm of an A.P. is given by, So, $164=S_{n}-S_{n-1}$ $164=3 n^{2}+5 n-\left[3(n-1)^{2}+5(n-1)\right]$ Using the property, $(a-b)^{2}=a^{2}+b^{2}-2 a b$ We get, $164=3 n^{2}+5 n-\left[3\left(n^{2}+1-2 n\right)...
Read More →if
Question: If $\tan ^{-1} x+\tan ^{-1} y=\frac{5 \pi}{6}$, then $\cot ^{-1} x+\cot ^{-1} y=$_________________. Solution: We know $\tan ^{-1} a+\cot ^{-1} a=\frac{\pi}{2}$, for all $a \in \mathrm{R}$ Now, $\tan ^{-1} x+\tan ^{-1} y=\frac{5 \pi}{6}$ (Given) $\Rightarrow \frac{\pi}{2}-\cot ^{-1} x+\frac{\pi}{2}-\cot ^{-1} y=\frac{5 \pi}{6}$ [Using (1)] $\Rightarrow \cot ^{-1} x+\cot ^{-1} y=\pi-\frac{5 \pi}{6}$ $\Rightarrow \cot ^{-1} x+\cot ^{-1} y=\frac{\pi}{6}$ If $\tan ^{-1} x+\tan ^{-1} y=\frac...
Read More →Write the total number of terms in the expansion of
Question: Write the total number of terms in the expansion of $(x+a)^{100}+(x-a)^{100}$. Solution: The total number of terms are 101 of which 50 terms get cancelled. Hence, the total number of terms in the expansion of $(x+a)^{100}+(x-a)^{100}$ is 51 ....
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