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- Question 5, Exercise 10.1
- 5(i)===== If $\tan \alpha =\dfrac{3}{4}$, $\sec \beta =\dfrac{13}{5}$ and neither the terminal side of the angle of measure $\alpha$ nor $\beta$ in the first Quadrant, then find: $\sin \left( \alpha +\beta \right)$. ====Solution==== Given: $\tan\alpha... quadrant, therefor it lies in 3rd quadrant. Now \begin{align}{{\sec}^{2}}\alpha &=1+{{\tan}^{2}}\alph
- Question11 and 12, Exercise 10.1
- r, Pakistan. =====Question 11===== If $\alpha$, $\beta$, $\gamma$ are the angles of a triangle $ABC$, show that $\cot \dfrac{\alpha }{2}+\cot \dfrac{\beta }{2}+\cot \dfrac{\gamma }{2}=\cot \dfrac{\alpha }{2}\cot \dfrac{\beta }{2}\cot \dfrac{\gamma }{2}$ ====Solution==== Since $\alpha$, $\beta$ and $\gamma$ are angles of triangle, therefore
- Question, Exercise 10.1
- ===== If $\sin \alpha =-\dfrac{4}{5}$ and $\cos \beta =-\dfrac{12}{13}$, $\alpha $in Quadrant III and $\beta $in Quadrant II, find the exact value of $\sin \left( \alpha -\beta \right)$. ====Solution==== Given: $\sin \alpha... c{4}{5}$, $\alpha$ is in 3rd quadrant, \\ $\sin \beta=-\dfrac{12}{13}$, $\beta$ is in 2nd quadrant. W
- Question 7, Exercise 10.2
- eta =\dfrac{1}{\sec 2\theta }$. ====Solution==== \begin{align}L.H.S&={{\cos }^{4}}\theta -{{\sin }^{4}... }{2}=\dfrac{2}{\sin \theta }$. ====Solution==== \begin{align}L.H.S&=\tan \dfrac{\theta }{2}+co\operat... \theta }{2}\cos \dfrac{\theta }{2}} \quad \because \left( {{\sin }^{2}}\theta +{{\cos }^{2}}\th... \theta }={{\cot }^{2}}\theta $. ====Solution==== \begin{align}L.H.S&=\dfrac{1+\cos 2\theta }{1+\cos 2\
- Question 7, Exercise 10.1
- tion 7(i)===== Show that: $\cot \left( \alpha +\beta \right)=\dfrac{\cot \alpha \cot \beta -1}{\cot \alpha +\cot \beta }$ ====Solution==== \begin{align}L.H.S.&=\cot (\alpha +\beta )\\ &=\dfrac{1}{\tan (\alpha +\beta )
- Question 1, Exercise 10.1
- }}\sin {{22}^{\circ }}$ ==== Solution ==== As \begin{align} \sin (\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta, \end{align} Therefore \begin{align} \sin {{37}^{\circ }}\cos {{22}^{\cir
- Question 8, Exercise 10.1
- }{\cos \theta -\sin \theta }$ ====Solution==== \begin{align}L.H.S.&=\tan \left( \dfrac{\pi }{4}+\the... {1-tan\theta }{1+tan\theta }$ ====Solution==== \begin{align}L.H.S.&=\tan \left( \dfrac{\pi }{4}-\the... \ &=R.H.S.\end{align} ===Alternative Method=== \begin{align}L.H.S.&=\tan\left( \dfrac{\pi }{4}-\thet... =\dfrac{1-\tan\theta}{1+1\cdot\tan\theta } \quad \because \tan\dfrac{\pi}{4}=1\\ &=\dfrac{\left( 1-\t
- Question 6, Exercise 10.1
- \sin }^{2}}\dfrac{\alpha }{2}$ ====Solution==== \begin{align}L.H.S&=\cos \alpha \\ \cos \alpha &=\cos... tion 6(ii)===== Show that: $\sin \left( \alpha +\beta \right)\sin \left( \alpha -\beta \right)={{\cos }^{2}}\beta -{{\cos }^{2}}\alpha$ ====Solution==== \begin{align}L.H.S.&=\sin \left(
- Question 1, Exercise 10.3
- n==== We have an identity: $$-2\sin \alpha \sin \beta =\cos (\alpha +\beta )-\cos (\alpha -\beta ).$$ Put $\alpha =6x$ and $\beta =x$ \begin{align}-\,2\sin 6x\sin x&=\cos (6x+x)-\cos (6x-x)\\ &
- Question 2, Exercise 10.3
- on==== We have an identity: $$\sin \alpha +\sin \beta =2\sin \left( \dfrac{\alpha +\beta }{2} \right)\cos \left( \dfrac{\alpha -\beta }{2} \right).$$ Put $\alpha ={{37}^{\circ }}$, $\beta ={{43}^{\circ }}$ \begin{align}\sin {{37}^{\cir
- Question 5, Exercise 10.3
- ==Solution==== We know that\\ $2\cos \alpha \cos \beta =\cos \left( \alpha +\beta \right)+\cos \left( \alpha -\beta \right)$\\ \begin{align}L.H.S.&=\cos {{20}^{\circ }}\cos {{40}^{\circ }}\cos {{60}^{\circ }}\cos {
- Question 3, Exercise 10.1
- and $\sin v=\dfrac{4}{5}$ where$u$ and $v$ are between $0$ and $\dfrac{\pi }{2}$, evaluate each of ... uadrant and $\cos $ is $+ve$ in first quadrant . \begin{align}\Rightarrow \,\,\,\,\,\cos u&=\sqrt{1-{{... quadrant and $\cos $ is $+ve$ in first quadrant. \begin{align}\Rightarrow \,\,\,\,\,\cos v&=\sqrt{1-{{... rrow \,\,\,\cos v &=\frac{3}{5}\end{align} Now \begin{align}\cos \left( u+v \right)&=\cos u\cos v-\s
- Question 13, Exercise 10.1
- phi$ and $3=r\sin\varphi$. Squaring and adding \begin{align} &r^2\cos^2 \varphi+r^2\sin^2\varphi = 4... lies &r^2 = 25\\ \implies &r=5 \end{align} Also \begin{align}r\cos\varphi = 4 \implies 5\cos\varphi =... implies \cos\varphi =\dfrac{4}{5}\end{align} and \begin{align}r\sin\varphi = 3 \implies 5\sin\varphi =... \dfrac{3}{5}.\end{align} Thus, from (1), we have \begin{align}&4\sin\theta +3\cos\theta \\ &=5 \left(\
- Question 2, Exercise 10.1
- pi }{3}-\dfrac{\pi }{4}$ and using the identity: \begin{align}\sin (\alpha -\beta )=\sin \alpha \cos \beta -\cos \alpha \sin.\end{align} \begin{align} \Rightarrow \quad \sin \left( \frac{\pi }{3}-\frac{\pi
- Question 5, Exercise 10.3
- ==Solution==== We know that\\ $2\cos \alpha \cos \beta =\cos \left( \alpha +\beta \right)+\cos \left( \alpha -\beta \right)$\\ \begin{align}L.H.S.&=\cos {{20}^{\circ }}\cos {{40}^{\circ }}\cos {{60}^{\circ }}\cos {