If cos (θ + ϕ) = m cos (θ – ϕ),
Question: If cos ( +ϕ) = m cos ( ϕ), then prove that tan = ((1 m)/(1 + m)) cot ϕ[Hint: Express cos ( + ϕ)/ cos ( ϕ) = m/l and apply Componendo and Dividendo] Solution: According to the question, $\cos (\theta+\phi)=m \cos (\theta-\phi)$ $\because \cos (\theta+\phi)=\mathrm{m} \cos (\theta-\phi)$ $\Rightarrow \frac{\cos (\theta-\phi)}{\cos (\theta+\phi)}=\frac{1}{m}$ Applying componendo - dividend, we get, $\Rightarrow \frac{\cos (\theta-\phi)+\cos (\theta+\phi)}{\cos (\theta-\phi)-\cos (\theta+\...
Read More →If sin (θ + α) = a and sin (θ + β) = b,
Question: If sin ( + ) = a and sin ( + ) = b, then prove that cos 2( ) 4ab cos ( ) = 1 2a2 2b2 Solution: According to the question, sin ( + ) = a and sin( + ) = b LHS = cos 2( ) 4ab cos ( ) Using cos 2x = 2cos2x 1, Let us solve, ⇒LHS = 2cos2( ) 1 4ab cos( ) ⇒LHS = 2cos ( ) {cos ( ) 2ab} 1 Since, cos ( ) = cos {( + ) ( + )} cos (A B) = cos A cos B + sin A sin B ⇒cos ( ) = cos( + )cos( + ) + sin( + )sin( + ) Since, sin( + ) = a ⇒cos( + ) = (1 sin2( + ) = (1 a2) Similarly, cos( + ) = (1 b2) Therefo...
Read More →If sec x cos 5x + 1 = 0,
Question: If sec x cos 5x + 1 = 0, where 0 x /2, then find the value of x. Solution: According to the question, sec x cos 5x = -1 ⇒cos 5x = -1/sec x We know that, sec x = 1/cos x ⇒cos 5x + cos x = 0 By transformation formula of T-ratios, We know that, $\cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right)$ $\Rightarrow 2 \cos \left(\frac{5 x+x}{2}\right) \cos \left(\frac{5 x-x}{2}\right)=0$ ⇒2 cos 3x cos 2x = 0 ⇒cos 3x = 0 or cos 2x = 0 ∵0 x /2 Therefore,0 2x or 0 3x 3/...
Read More →Prove that the function
Question: Prove that the function $f(x)=x^{3}-6 x^{2}+12 x-18$ is increasing on $R$. Solution: Given:- Function $f(x)=x^{3}-6 x^{2}+12 x-18$ Theorem:- Let $f$ be a differentiable real function defined on an open interval $(a, b)$. (i) If $f^{\prime}(x)0$ for all $x \in(a, b)$, then $f(x)$ is increasing on $(a, b)$ (ii) If $f^{\prime}(x)0$ for all $x \in(a, b)$, then $f(x)$ is decreasing on $(a, b)$ Algorithm:- (i) Obtain the function and put it equal to $f(x)$ (ii) Find $f^{\prime}(x)$ (iii) Put...
Read More →If 2sin2θ = 3cos θ,
Question: If 2sin2 = 3cos , where 0 2, then find the value of . Solution: According to the question, 2sin2 = 3cos We know that, sin2 = 1 cos2 Given that, 2 sin2 = 3 cos 2 2 cos2 = 3 cos 2 cos2 + 3 cos 2= 0 (cos + 2)(2 cos 1) = 0 Therefore, cos = = cos /3 = /3 or 2 /3 = /3, 5/3 Therefore,2(1 cos2) = 3cos ⇒2 2cos2 = 3cos ⇒2cos2 + 3cos 2 = 0 ⇒2cos2 + 4cos cos 2 = 0 ⇒2cos (cos + 2) +1 (cos + 2) = 0 ⇒(2cos + 1)(cos + 2) = 0 Since, cos [-1,1] , for any value . So, cos 2 Therefore, 2 cos 1 = 0 ⇒cos = =...
Read More →If cot θ + tan θ = 2 cosec θ,
Question: If cot + tan = 2 cosec , then find the general value of . Solution: According to the question, $\Rightarrow \frac{\cos \theta}{\sin \theta}+\frac{\sin \theta}{\cos \theta}=2 \operatorname{cosec} \theta$ Since, $\sin ^{2} \theta+\cos ^{2} \theta=1$ $\Rightarrow \frac{\cos ^{2} \theta+\sin ^{2} \theta}{\sin \theta \cos \theta}=2 \operatorname{cosec} \theta$ ⇒1 = 2 cosec sin cos We know that, sin cosec = 1 ⇒1 = 2 cos ⇒cos = 1/2 = cos(/3) Hence, Thesolution of cos x = cos can be given by, ...
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Question: Show that $f(x)=x^{9}+4 x^{7}+11$ is an increasing function for all $x \in R$. Solution: Given:- Function $f(x)=x^{9}+4 x^{7}+11$ Theorem:- Let $f$ be a differentiable real function defined on an open interval $(a, b)$. (i) If $f^{\prime}(x)0$ for $a l l_{x} \in(a, b)$, then $f(x)$ is increasing on $(a, b)$ (ii) If $f^{\prime}(x)0$ for all $x \in(a, b)$, then $f(x)$ is decreasing on $(a, b)$ Algorithm:- (i) Obtain the function and put it equal to $f(x)$ (ii) Find $f^{\prime}(x)$ (iii) ...
Read More →Find the most general value of θ
Question: Find the most general value of satisfying the equation tan = 1 and cos = 1/2 Solution: According to the question, We have, tan = -1 And cos =1/2 . ⇒ = /4 So, we know that, lies in IV quadrant. = 2 /4 = 7/4 So, general solution is = 7/4 + 2 n , n Z...
Read More →Show that the function
Question: Show that the function $x^{2}-x+1$ is neither increasing nor decreasing on $(0,1)$. Solution: Given:- Function $f(x)=x^{2}-x+1$ Theorem:- Let $f$ be a differentiable real function defined on an open interval $(a, b)$. (i) If $f^{\prime}(x)0$ for all $x \in(a, b)$, then $f(x)$ is increasing on $(a, b)$ (ii) If $f^{\prime}(x)0$ for all $x \in(a, b)$, then $f(x)$ is decreasing on $(a, b)$ Algorithm:- (i) Obtain the function and put it equal to $f(x)$ (ii) Find $f^{\prime}(x)$ (iii) Put $f...
Read More →Find the image of the point
Question: Find the image of the point P(1, 2) in the line x 3y + 4 = 0. Solution: Let line AB be x 3y + 4 = 0 and point P be (1, 2) Let the image of the point P(1, 2) in the line mirror AB be Q(h, k) Since line $A B$ is a mirror. Then $P Q$ is perpendicularly bisected at $R$. Since $R$ is the midpoint of $P Q$. We know that, Midpoint of a line joining $\left(x_{1}, y_{1}\right) \\left(x_{2}, y_{2}\right)=\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}$ So, Midpoint of the line joining $(1,2) \(h, k...
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Question: Show that $f(x)=(x-1) e^{x}+1$ is an increasing function for all $x0$. Solution: Given:- Function $f(x)=(x-1) e^{x}+1$ Theorem:- Let $f$ be a differentiable real function defined on an open interval $(a, b)$. (i) If $f^{\prime}(x)0$ for $a l l x \in(a, b)$, then $f(x)$ is increasing on $(a, b)$ (ii) If $f^{\prime}(x)0$ for $a l l x \in(a, b)$, then $f(x)$ is decreasing on $(a, b)$ Algorithm:- (i) Obtain the function and put it equal to $f(x)$ (ii) Find $f^{\prime}(x)$ (iii) Put $f^{\pr...
Read More →If sin θ + cos θ = 1,
Question: If sin + cos = 1, then find the general value of . Solution: According to the question, sin + cos = 1 As, sin + cos = 1 $\Rightarrow \sqrt{2}\left(\frac{1}{\sqrt{2}} \sin \theta+\frac{1}{\sqrt{2}} \cos \theta\right)=1$ We know that, $\sin (\pi / 4)=\cos (\pi / 4)=1 / \sqrt{2}$ $\Rightarrow \sqrt{2}\left(\sin \theta \cos \frac{\pi}{4}+\sin \frac{\pi}{4} \cos \theta\right)=1$ We know that $\sin (A+B)=\sin A \cos B+\cos A \sin B$ $\Rightarrow \sin \left(\frac{\pi}{4}+\theta\right)=\frac{1...
Read More →Prove the following
Question: If $\tan \theta=\frac{\sin \alpha-\cos \alpha}{\sin \alpha+\cos \alpha}$, then show that $\sin \alpha+\cos \alpha=\sqrt{2} \cos \theta$ [Hint: Express tan = tan( /2) = /4] Solution: We know that, $\tan \theta=\frac{\sin \alpha-\cos \alpha}{\sin \alpha+\cos \alpha}$ $\Rightarrow \tan \theta=\frac{\cos \alpha\left(\frac{\sin \alpha}{\cos \alpha}-1\right)}{\cos \alpha\left(\frac{\sin \alpha}{\cos \alpha}+1\right)}$ Since, $\tan \mathrm{A}=(\sin \mathrm{A}) /(\cos \mathrm{A})$ $\Rightarrow...
Read More →Show that the function
Question: Show that the function $f(x)=\cot ^{-1}(\sin x+\cos x)$ is decreasing on $(0, \pi / 4)$ and increasing on $(\pi / 4, \pi / 2)$. Solution: Given:- Function $f(x)=\cot ^{-1}(\sin x+\cos x)$ Theorem:- Let $f$ be a differentiable real function defined on an open interval $(a, b)$. (i) If $f^{\prime}(x)0$ for all $x \in(a, b)$, then $f(x)$ is increasing on $(a, b)$ (ii) If $f^{\prime}(x)0$ for all $x \in(a, b)$, then $f(x)$ is decreasing on $(a, b)$ Algorithm:- (i) Obtain the function and p...
Read More →Prove the following
Question: If $\frac{\sin (x+y)}{\sin (x-y)}=\frac{a+b}{a-b}$, then show that $\frac{\tan x}{\tan y}=\frac{a}{b}$. [Hint: Use componendo and Dividendo] Solution: According to the question, $\frac{\sin (x+y)}{\sin (x-y)}=\frac{a+b}{a-b}$ Since, $\sin (A+B)=\sin A \cos B+\cos A \sin B$ $\therefore \frac{\sin (x+y)}{\sin (x-y)}=\frac{a+b}{a-b}$ $\Rightarrow \frac{\sin x \cos y+\cos x \sin y}{\sin x \cos y-\cos x \sin y}=\frac{a+b}{a-b}$ Applying componendo-dividendo rule, We get, $\Rightarrow \frac{...
Read More →If cosα + cosβ = 0 = sinα + sinβ,
Question: If cos + cos = 0 = sin + sin, then prove that cos 2 + cos 2 = 2cos ( + ).[Hint: cos + cos)2 (sin + sin)2 = 0] Solution: According to the question, cos + cos = 0 = sin + sin (i) Since,LHS = cos 2 + cos 2 We know that, cos 2x = cos2x sin2x Therefore, LHS = cos2 sin2 + (cos2 sin2) ⇒LHS = cos2 + cos2 (sin2 + sin2) Also, since, a2+ b2= (a+b)2 2ab ⇒LHS = (cos + cos)2 2cos cos (sin + sin)2+2sin sin From equation (i), ⇒LHS = 0 2cos cos -0 + 2sin sin ⇒LHS = -2(cos cos sin sin) ∵cos ( + ) = cos ...
Read More →If tan (A + B) = p, tan (A – B) = q,
Question: If tan (A + B) = p, tan (A B) = q, then show thattan 2A = (p + q) / (1 pq).[Hint: Use 2A = (A + B) + (A B)] Solution: We know that, tan 2A = tan (A + B + A B) And also, $\tan (\mathrm{x}+\mathrm{y})=\frac{\tan \mathrm{x}+\tan \mathrm{y}}{1-\tan \mathrm{x} \tan y}$ $\therefore \tan 2 \mathrm{~A}=\frac{\tan (\mathrm{A}+\mathrm{B})+\tan (\mathrm{A}-\mathrm{B})}{1-\tan (\mathrm{A}+\mathrm{B}) \tan (\mathrm{A}-\mathrm{B})}$ Substituting the values given in question, $\Rightarrow \tan 2 A=\f...
Read More →Find the value of k so that the lines
Question: Find the value of k so that the lines $3 x-y-2=0,5 x+k y-3=0$ and $2 x+y-3 = 0 are concurrent. Solution: Given that $3 x-y-2=0$, $5 x+k y-3=0$ and $2 x+y-3=0$ are concurrent We know that, The lines $a_{1} x+b_{1} y+c_{1}=0, a_{1} x+b_{1} y+c_{1}=0$ and $a_{1} x+b_{1} y+c_{1}=0$ are concurrent if It is given that the given lines are concurrent. Now, expanding along first row, we get $\Rightarrow 3[(k)(-3)-(-3)(1)]-(-1)[(5)(-3)-(-3)(2)]+(-2)[5-2 k]=0$ $\Rightarrow 3[-3 k+3]+1[-15+6]-2[5-...
Read More →Show that the function
Question: Show that the function $\mathrm{f}(\mathrm{x})=\sin \left(2 \mathrm{x}+\frac{\pi}{4}\right)$ is decreasing on $\left(\frac{3 \pi}{8}, \frac{5 \pi}{8}\right)$ Solution: Given:- Function $\mathrm{f}(\mathrm{x})=\sin \left(2 \mathrm{x}+\frac{\pi}{4}\right)$ Theorem:- Let $\mathrm{f}$ be a differentiable real function defined on an open interval $(\mathrm{a}, \mathrm{b})$. (i) If $f^{\prime}(x)0$ for $a l l x \in(a, b)$, then $f(x)$ is increasing on $(a, b)$ (ii) If $f^{\prime}(x)0$ for al...
Read More →If tan θ + sin θ = m and tan θ – sin θ = n,
Question: If tan + sin = m and tan sin = n, then prove that m2 n2= 4 sin tan [Hint: m + n = 2tan, m n = 2 sin , then use m2 n2 = (m + n)(m n)] Solution: According to the question, tan + sin = m (i) tan sin = n (ii) Adding equation i and ii, 2 tan = m + n (iii) Subtracting equation ii from i, We get, 2sin = m n (iv) Multiplying equations (iii) and (iv), 2sin (2tan ) = (m + n)(m n) ⇒4 sin tan = m2 n2 Hence, m2 n2 = 4 sin tan...
Read More →Prove that sin 4A = 4sinA cos3A – 4 cosA sin3A.
Question: Prove that sin 4A = 4sinA cos3A 4 cosA sin3A. Solution: sin 4A = sin (2A + 2A) We know that, sin(A + B) = sin A cos B + cos A sin B Therefore,sin 4A = sin 2A cos 2A + cos 2A sin 2A ⇒sin 4A = 2 sin 2A cos 2A From T-ratios of multiple angle, We get, sin 2A = 2 sin A cos A and cos 2A = cos2A sin2A ⇒sin 4A = 2(2 sin A cos A)(cos2A sin2A) ⇒sin 4A = 4 sin A cos3A 4 cos A sin3A Hence, sin 4A = 4 sin A cos3A 4 cos A sin3A...
Read More →Find the value of tan 22°30’.
Question: Find the value of tan 2230. [Hint: Let $\theta=45^{\circ}$, use $\left.\tan \frac{\theta}{2}=\frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}=\frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2}}{2 \cos ^{2} \frac{\theta}{2}}=\frac{\sin \theta}{1+\cos \theta}\right]$ Solution: Let, = 45 As we need to find: tan 2230 = tan (/2) We know that, sin = cos = 1/2 (for = 45) Since, $\tan \frac{\theta}{2}=\frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}$ Multiplying $2 \cos \theta / 2$ in n...
Read More →Show that
Question: Show that $f(x)=\tan ^{-1}(\sin x+\cos x)$ is a decreasing function on the interval $(\pi / 4, \pi / 2)$. Solution: Given:- Function $f(x)=\tan ^{-1}(\sin x+\cos x)$ Theorem:- Let $f$ be a differentiable real function defined on an open interval $(a, b)$. (i) If $f^{\prime}(x)0$ for all $x \in(a, b)$, then $f(x)$ is increasing on $(a, b)$ (ii) If $f^{\prime}(x)0$ for all $x \in(a, b)$, then $f(x)$ is decreasing on $(a, b)$ Algorithm:- (i) Obtain the function and put it equal to $f(x)$ ...
Read More →If a cos θ + b sin θ = m and a sin θ – b cos θ = n,
Question: If a cos + b sin = m and a sin b cos = n, then show that a2+ b2= m2+ n2. Solution: According to the question, a cos + b sin = m (i) a sin b cos = n (ii) Squaring and adding equation 1 and 2, we get, (a cos + b sin )2+ (a sin b cos )2= m2+ n2 ⇒a2cos2 + b2sin2 + 2ab sin cos + a2sin2 + b2cos2 2ab sin cos = m2+ n2 ⇒a2cos2 + b2sin2 + a2sin2 + b2cos2 = m2+ n2 ⇒a2(sin2 + cos2) + b2(sin2 + cos2) = m2+ n2 Using, sin2 + cos2 = 1, We get, ⇒a2+ b2= m2+ n2...
Read More →Prove that cos θ cos θ/2 – cos 3θ cos 9θ/2 = sin7θ sin4θ
Question: Prove that cos cos /2 cos 3 cos 9/2 = sin7 sin4[Hint: Express L.H.S. = [2cos cos /2 2cos 3 cos 9 / 2] Solution: Using transformation formula, we get, 2 cos A cos B = cos(A + B) + cos (A B) -2 sin A sin B = cos(A + B) cos (A B) Multiplying and dividing the expression by 2. $\therefore$ LHS $=\frac{1}{2}\left(2 \cos \theta \cos \frac{\theta}{2}-2 \cos 3 \theta \cos \frac{9 \theta}{2}\right)$ Applying transformation formula, we get, LHS $=\frac{1}{2}\left(\cos \left(\theta+\frac{\theta}{2...
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