If $\cos ^{-1} \frac{x}{2}+\cos ^{-1} \frac{y}{3}=\theta$, then $9 x^{2}-12 x y \cos \theta+4 y^{2}$ is equal to
[question] Question. If $\cos ^{-1} \frac{x}{2}+\cos ^{-1} \frac{y}{3}=\theta$, then $9 x^{2}-12 x y \cos \theta+4 y^{2}$ is equal to (a) 36 (b) $-36 \sin ^{2} \theta$ (c) $36 \sin ^{2} \theta$ (d) $36 \cos ^{2} \theta$ [/question] [solution] Solution: (c) $36 \sin ^{2} \theta$ We know $\cos ^{-1} x+\cos ^{-1} y=\cos ^{-1}\left[x y-\sqrt{1-x^{2}} \sqrt{1-y^{2}}\right]$ $\Rightarrow \cos ^{-1}\left[\frac{x}{2} \frac{y}{3}-\sqrt{1-\frac{x^{2}}{4}} \sqrt{1-\frac{y^{2}} {3}}\right]=\theta$ $\Rightar...
Read More →If (m + 1)th term of an A.P. is twice the (n + 1)th term,
Question: If (m+ 1)th term of an A.P. is twice the (n+ 1)th term, prove that (3m+ 1)th term is twice the (m+n+ 1)th term. Solution: Given: $a_{m+1}=2 a_{n+1}$ $\Rightarrow a+(m+1-1) d=2[a+(n+1-1) d]$ $\Rightarrow a+m d=2(a+n d)$ $\Rightarrow a+m d=2 a+2 n d$ $\Rightarrow 0=a+2 n d-m d$ $\Rightarrow n d=\frac{m d-a}{2} \quad \ldots(\mathrm{i})$ To prove: $a_{3 m+1}=2 a_{m+n+1}$ LHS : $a_{3 m+1}=a+(3 m+1-1) d$ $\Rightarrow a_{3 m+1}=a+3 m d$ RHS : $2 a_{m+n+1}=2[a+(m+n+1-1) d]$ $\Rightarrow 2 a_{m...
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Question: If $y=2 \tan ^{-1} x+\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)$ for all $x$, then $y$ lies in the interval_________________. Solution: We know $2 \tan ^{-1} x= \begin{cases}\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right), -1 \leq x \leq 1 \\ \pi-\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right), x1 \\ -\pi-\sin ^{-1}\left(\frac{2 \mathrm{x}}{1+\mathrm{x}^{2}}\right), x-1\end{cases}$ $\therefore y=2 \tan ^{-1} x+\sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)= \begin{cases}4 \tan ^{-1} x, -1 \leq x \le...
Read More →In ∆ABC, AB = AC and ∠B = 50°.
Question: In∆ABC,AB=ACand B= 50. Then, A= ?(a) 40(b) 50(c) 80(d) 130 Solution: In $\triangle A B C$, we have: $A B=A C$ $\angle B=50^{\circ}$ Since $A B C$ is an isosceles triangle, we have: $\angle C=\angle B$ $\angle C=50^{\circ}$ In triangle $\mathrm{ABC}$, we have: $\angle A+\angle B+\angle C=180^{\circ}$ $\Rightarrow \angle A+50+50=180^{\circ}$ $\Rightarrow \angle A=180^{\circ}-100^{\circ}$ $\therefore \angle A=80^{\circ}$ Hence, the correct answer is option (c)....
Read More →In the given figure, ABC is a right triangle right-angled at
Question: In the given figure, ABC is a right triangle right-angled at B such that BC = 6 cm and AB = 8 cm. Find the radius of its incircle. Solution: From the property of tangents we know that the length of two tangents drawn to a circle from the same external point will be equal. Therefore, we have BQ = BP Let us denoteBPandBQbyx AP = AR Let us denoteAPandARbyy RC = QC Let us denoteRCandRQbyz We have been given thatis a right triangle andBC= 6 cm andAB= 8 cm. let us find outACusing Pythagoras ...
Read More →If ∆ABC ≅ ∆PQR then which of the following is not true?
Question: If∆ABC ∆PQR then which of the following is not true?(a)BC=PQ(b)AC=PR(c)BC=QR(d)AB=PQ Solution: (a) BC = PQ If $\Delta A B C \cong \Delta P Q R$, then BC =QRHence, the correct answer is option (a)....
Read More →If AB = QR, BC = RP and CA = PQ, then which of the following holds?
Question: IfAB=QR,BC=RPandCA=PQ, then which of the following holds?(a)∆ABC ∆PQR(b)∆CBA ∆PQR(c)∆CAB ∆PQR(d)∆BCA ∆PQR Solution: (c) $\Delta C A B \cong \Delta P Q R$ As, $A B=Q R, \quad$ (given) $B C=R P, \quad$ (given) $\mathrm{CA}=\mathrm{PQ} \quad$ (given) $\therefore \Delta C A B \cong \Delta P Q R$...
Read More →From an external point P, tangents PA and PB are drawn to a circle with centre O.
Question: From an external pointP, tangentsPAandPBare drawn to a circle with centreO. IfCDis the tangent to the circle at a pointEandPA= 14 cm, find the perimeter of ΔPCD. Solution: Let us first put the given data in the form of a diagram. It is given that PA = 14cm. we have to find the perimeter of. Perimeter ofisPC + CD + PD Looking at the figure we can rewrite the equation as follows. Perimeter ofisPC + CE + ED + PD(1) From the property of tangents we know that the length of two tangents draw...
Read More →Which of the following is not a criterion for congruence of triangles?
Question: Which of the following is not a criterion for congruence of triangles?(a) SSA(b) SAS(c) ASA(d) SSS Solution: (a) SSASSA is not a criterion for congruence of triangles....
Read More →Solve this
Question: $\tan ^{-1}\left(\tan \frac{2 \pi}{3}\right)$ is equal to__________________. Solution: $\tan ^{-1}\left(\tan \frac{2 \pi}{3}\right)$ $=\tan ^{-1}\left[\tan \left(\pi-\frac{\pi}{3}\right)\right]$ $=\tan ^{-1}\left(-\tan \frac{\pi}{3}\right)$ $=\tan ^{-1}\left[\tan \left(-\frac{\pi}{3}\right)\right]$ $[\tan (-\theta)=-\tan \theta]$ $=-\frac{\pi}{3}$ $\left[\tan ^{-1}(\tan x)=x\right.$, if $\left.x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\right]$ $\tan ^{-1}\left(\tan \frac{2 \pi}{3}...
Read More →If PA and PB are tangents from an outside point P such
Question: IfPAandPBare tangents from an outside pointPsuch thatPA= 10 cm and APB= 60. Find the length of chordAB. Solution: Let us first put the given data in the form of a diagram. From the property of tangents we know that the length of two tangents drawn to a circle from a common external point will always be equal. Therefore, PA=PB Consider the trianglePAB. Since we havePA=PB, it is an isosceles triangle. We know that in an isosceles triangle, the angles opposite to the equal sides will be e...
Read More →In a certain A.P. the 24th term is twice the 10th term.
Question: In a certain A.P. the 24th term is twice the 10th term. Prove that the 72nd term is twice the 34th term. Solution: Given: $a_{24}=2 a_{10}$ $\Rightarrow a+(24-1) d=2[a+(10-1) d]$ $\Rightarrow a+23 d=2(a+9 d)$ $\Rightarrow a+23 d=2 a+18 d$ $\Rightarrow 5 d=a \quad \ldots$ (i) To prove: $a_{72}=2 a_{34}$ LHS : $a_{72}=a+(72-1) d$ $\Rightarrow a_{72}=a+71 d$ $\Rightarrow a_{72}=5 d+71 d \quad(\operatorname{From}(\mathrm{i}))$ $\Rightarrow a_{72}=76 d$ RHS : $2 a_{34}=2[a+(34-1) d]$ $\Righ...
Read More →In the given figure, D is a point on side BC of a ΔABC and E is a point such that CD = DE. Prove that AB + AC > BE.
Question: In the given figure,Dis a point on sideBCof a ΔABCandEis a point such thatCD=DE. Prove thatAB+ACBE. Solution: Given:CD=DETo prove:AB+ACBEProof: In $\Delta A B C$ $A B+A CB C \quad \ldots(1)$ In $\Delta B E D$ $B D+C DB E$ $\Rightarrow B CB E \quad \ldots(2)$ From (1) and (2), we get $A B+A CB E$...
Read More →In the given figure, PQ is tangent at a point R of the circle
Question: In the given figure,PQis tangent at a pointRof the circle with centreO. If TRQ= 30, findmPRS. Solution: We have been given that $\angle T R Q=30^{\circ}$. From the property of tangents we know that a tangent will always be perpendicular to the radius at the point of contact. Therefore, Looking at the given figure we can rewrite the above equation as follows, We know that. Therefore, $30^{\circ}+\angle T R O=90^{\circ}$ $\angle T R O=60^{\circ}$ Now consider. The two sides of this trian...
Read More →$\tan ^{-1} \frac{1}{11}+\tan ^{-1} \frac{2}{11}$ is equal to
[question] Question. $\tan ^{-1} \frac{1}{11}+\tan ^{-1} \frac{2}{11}$ is equal to (a) 0 (b) $1 / 2$ (c) $-1$ (d) none of these [/question] [solution] Solution: (d) none of these We know that $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$. Now, $\tan ^{-1} \frac{1}{11}+\tan ^{-1} \frac{2}{11}=\tan ^{-1}\left(\frac{\frac{1}{11}+\frac{2}{11}}{1-\frac{1} {11} \frac{2}{11}}\right)$ $=\tan ^{-1}\left(\frac{\frac{3}{11}}{\frac{121-2}{121}}\right)$ $=\tan ^{-1}\left(\frac{\frac{3}...
Read More →In a certain A.P. the 24th term is twice the 10th term.
Question: In a certain A.P. the 24th term is twice the 10th term. Prove that the 72nd term is twice the 34th term. Solution: Given: $a_{24}=2 a_{10}$ $\Rightarrow a+(24-1) d=2[a+(10-1) d]$ $\Rightarrow a+23 d=2(a+9 d)$ $\Rightarrow a+23 d=2 a+18 d$ $\Rightarrow 5 d=a \quad \ldots$ (i) To prove: $a_{72}=2 a_{34}$ LHS : $a_{72}=a+(72-1) d$ $\Rightarrow a_{72}=a+71 d$ $\Rightarrow a_{72}=5 d+71 d \quad(\operatorname{From}(\mathrm{i}))$ $\Rightarrow a_{72}=76 d$ RHS : $2 a_{34}=2[a+(34-1) d]$ $\Righ...
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Question: $\cot \left(\frac{\pi}{4}-2 \cot ^{-1} 3\right)$ is equal to______________________. Solution: Let $\cot ^{-1} 3=\theta \Rightarrow \cot \theta=3$ $\therefore \cot \left(\frac{\pi}{4}-2 \cot ^{-1} 3\right)$ $=\cot \left(\frac{\pi}{4}-2 \theta\right)$ $=\frac{\cot \frac{\pi}{4} \cot 2 \theta+1}{\cot 2 \theta-\cot \frac{\pi}{4}}$ $\left[\cot (A-B)=\frac{\cot A \cot B+1}{\cot B-\cot A}\right]$ $=\frac{\cot 2 \theta+1}{\cot 2 \theta-1}$ $=\frac{\frac{\cot ^{2} \theta-1}{2 \cot \theta}+1}{\f...
Read More →Prove that the intercept of a tangent between two parallel
Question: Prove that the intercept of a tangent between two parallel tangents to a circle subtends a right angle at the centre. Solution: Given:XYandXYat are two parallel tangents to the circle with centreOandABis the tangent at the pointC, which intersectsXYatAandXYat B. To Prove:AOB= 90 Construction: Let us joint pointOtoC. Proof: InΔOPAandΔOCA, we have OP = OC(Radii of the same circle) AP = AC(Tangents from pointA) AO = AO(Common side) ΔOPAΔOCA(SSS congruence criterion) Therefore,POA=COA(i) (...
Read More →If AB, AC, PQ are tangents in the given figure and AB = 5 cm,
Question: If AB, AC, PQ are tangents in the given figure and AB = 5 cm, find the perimeter of Δ APQ. Solution: We have been asked to find the perimeter of the triangleAPQ. Therefore, Perimeter of ΔAPQis equal toAP + AQ + PQ By looking at the figure, we can rewrite the above as follows, Let the Perimeter of ΔAPQbeP. SoP=AP + AQ + PX + XQ From the property of tangents we know that when two tangents are drawn to a circle from the same external point, the length of the two tangents will be equal. Th...
Read More →The 10th and 18th terms of an A.P. are 41 and 73 respectively.
Question: The 10th and 18th terms of an A.P. are 41 and 73 respectively. Find 26th term. Solution: Given: $a_{10}=41$ $\Rightarrow a+(10-1) d=41 \quad\left[a_{n}=a+(n-1) d\right]$ $\Rightarrow a+9 d=41$ And, $a_{18}=73$ $\Rightarrow a+(18-1) d=73 \quad\left[a_{n}=a+(n-1) d\right]$ $\Rightarrow a+17 d=73$ Solving the two equations, we get: $\Rightarrow 17 d-9 d=73-41$ $\Rightarrow 8 d=32$ $\Rightarrow d=4$ ...(i) Putting the value in first equation, we get: $a+9 \times 4=41$ $\Rightarrow a+36=41$...
Read More →In the given figure, AD ⊥ BC and CD > BD. Show that AC > AB.
Question: In the given figure,ADBCandCDBD. Show thatACAB. Solution: Given:ADBCandCDBDTo prove:ACABProof: $\angle A D B=\angle A D C=90^{\circ} \quad(A D \perp B C) \quad \ldots(1)$ $\angle B A D\angle D A C \quad(C DB D) \quad \ldots(2)$ In $\Delta A B D$ Using angle sum property of a triangle, $\angle B=180^{\circ}-\angle A D B-\angle B A D$ $\angle B=90^{\circ}-\angle B A D \quad \ldots(3)$ In $\Delta A D C$ Using angle sum property of a triangle, $\angle A C D=90^{\circ}-\angle D A C \quad \l...
Read More →Let $f(x)=e^{\cos ^{-1}\{\sin (x+\pi / 3)\}}$. Then, $f(8 \pi / 9)=$
[question] Question. Let $f(x)=e^{\cos ^{-1}\{\sin (x+\pi / 3)\}}$. Then, $f(8 \pi / 9)=$ (a) $e^{5 \pi / 18}$ (b) $e^{13 \pi / 18}$ (c) $e^{-2 \pi / 18}$ (d) none of these [question] [solution] Solution: (b) $e^{13 \pi / 18}$ Given: $f(x)=e^{\cos ^{-1}\{\sin (x+\pi / 3)\}}$ Then, $f(8 \pi / 9)=e^{\cos ^{-1}\{\sin (8 \pi / 9+\pi / 3)\}}$ $=e^{\cos ^{-1}\{\sin (11 \pi / 9)\}}$ $=e^{\cos ^{-1}\left\{\cos \left(\pi / 2+{ }^{13 \pi} / 18\right)\right\}} \quad[\because \cos (\pi / 2+\theta)=\sin \the...
Read More →If the sides of a quadrilateral touch a circle.
Question: If the sides of a quadrilateral touch a circle. prove that the sum of a pair of opposite sides is equal to the sum of the other pair. Solution: Let us first put the given data in the form of a diagram. We have been asked to prove that the sum of the pair of opposite sides of the quadrilateral is equal to the sum of the other pair. Therefore, we shall first consider, AB + DC But by looking at the figure we have, AB + DC = AF + FB + DH + HC (1) From the property of tangents we know that ...
Read More →if
Question: If $\tan ^{-1} x+\tan ^{-1} \frac{1}{2}=\frac{\pi}{4}$, then $x=$_________________. Solution: $\tan ^{-1} x+\tan ^{-1} \frac{1}{2}=\frac{\pi}{4}$ $\Rightarrow \tan ^{-1}\left(\frac{x+\frac{1}{2}}{1-x \times \frac{1}{2}}\right)=\frac{\pi}{4}$ $\left[\tan ^{-1} a+\tan ^{-1} b=\tan ^{-1}\left(\frac{a+b}{1-a b}\right)\right]$ $\Rightarrow \frac{2 x+1}{2-x}=\tan \frac{\pi}{4}=1$ $\Rightarrow 2 x+1=2-x$ $\Rightarrow 3 x=1$ $\Rightarrow x=\frac{1}{3}$ Thus, the value of $x$ is $\frac{1}{3}$. ...
Read More →If from any point on the common chord of two interesting circle,
Question: If from any point on the common chord of two interesting circle, tangents be drawn to the circles, prove that they are equal. Solution: Let the two circles intersect at points X and Y. XY is the common chord. Suppose A is a point on the common chord and AM and AN be the tangents drawn from A to the circle. We need to show that AM = AN. In order to prove the above relation, following property will be used. Let PT be a tangent to the circle from an external point P and a secant to the ci...
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