Exercise 8.1
Question 31
यदि 2tanθ=1 , तब $\frac{3 \cos \theta+\sin \theta}{2 \cos \theta-\sin \theta}$ का मूल्य ज्ञात करे।
Sol :
tan θ=लम्ब/आधार$=\frac{1}{2}=\frac{AB}{BC}$
$AC=\sqrt{(AB)^{2}-(BC)^{2}}$
$AC=\sqrt{(1)^{2}-(2)^{2}}=\sqrt{1+4}$
$AC=\sqrt{5}$
sin θ=लम्ब/कर्ण$=\frac{AB}{AC}=\frac{1}{\sqrt{5}}$
cos θ=आधार/कर्ण $=\frac{BC}{AC}=\frac{2}{\sqrt{5}}$
$\frac{3 \cos \theta+\sin }{2 \cos \theta-\sin \theta}=\frac{2 \times \frac{2}{\sqrt{5}}+\frac{1}{\sqrt{5}}}{2 \times \frac{2}{\sqrt{5}}-\frac{1}{\sqrt{5}}}$
$=\frac{\frac{6}{\sqrt{5}}+\frac{1}{\sqrt{5}}}{\frac{4}{\sqrt{5}}-\frac{1}{\sqrt{5}}}=\frac{\frac{6+1}{\sqrt{5}}}{\frac{2-1}{\sqrt{5}}}$
$=\frac{\frac{7}{\sqrt{5}}}{\frac{1}{\sqrt{5}}}=\frac{7}{3}$
Question 32
यदि 5tan⍺=4 , तो दिखायें कि $\frac{5 \sin \alpha-3 \cos \alpha}{5 \sin \alpha+2 \cos \alpha}=\frac{1}{6}$
Sol :
$\mathrm{AC}=\sqrt{(a b)^{2}+(b c)^{2}}$
$\mathrm{AC}=\sqrt{(4)^{2}+(5)^{2}}=\sqrt{16+25}$
$\mathrm{AC}=\sqrt{41}$
sin ⍺=लम्ब/कर्ण$=\frac{AB}{AC}=\frac{4}{\sqrt{41}}$
cos ⍺=आधार/कर्ण$=\frac{BC}{AC}=\frac{5}{\sqrt{41}}$
LHS
$\frac{5 \sin \alpha-3 \cos \alpha}{5 \sin \alpha+2 \cos \alpha}=\frac{5 \times \frac{4}{\sqrt{41}}-3 \times \frac{5}{\sqrt{41}}}{5 \times \frac{4}{\sqrt{41}}+2 \times \frac{5}{\sqrt{41}}}$
$=\frac{\frac{20}{\sqrt{41}}-\frac{15}{\sqrt{41}}}{\frac{20}{\sqrt{41}}+\frac{10}{\sqrt{41}}}=\frac{\frac{20-15}{\sqrt{41}}}{\frac{20+10}{\sqrt{41}}}$
$=\frac{5}{30}=\frac{1}{6}$
LHS=RHLS proved
Question 33
यदि $\cot \theta=\frac{3}{4}$ , तो सिद्ध करें कि $\sqrt{\frac{\sec \theta+\operatorname{cosec} \theta}{\sec \theta-\operatorname{cosec} \theta}}=\sqrt{7}$
Sol :
cot θ=आधार/लम्ब$=\frac{3}{4}=\frac{BC}{AB}$
$\mathrm{AC}=\sqrt{(a b)^{2}+(b c)^{2}}$
$\mathrm{AC}=\sqrt{(4)^{2}+(3)^{2}}=\sqrt{16+9}$
$\mathrm{AC}=\sqrt{5}$
sec θ=कर्ण/आधार$=\frac{AC}{BC}=\frac{5}{3}$
cosec θ=$=\frac{AC}{AB}=\frac{5}{4}$
LHS
$\sqrt{\frac{\sec \theta-\operatorname{coses} \theta}{\sec \theta-\operatorname{cosec} \theta}}=\sqrt{\frac{\frac{5}{3}+\frac{5}{4}}{\frac{5}{3}-\frac{5}{4}}}$
$=\sqrt{\frac{\frac{20+15}{12}}{\frac{20-15}{12}}}=\sqrt{\frac{\frac{35}{12}}{\frac{5}{12}}}$
$=\sqrt{\frac{35}{5}}=\sqrt{7}$
LHS=RHS proved
Question 34
यदि $\cot \theta=\frac{1}{\sqrt{3}}$ , तब सत्यापित करें कि $\frac{1-\cos ^{2}\theta}{2-\sin ^{2} \theta}=\frac{3}{5}$
Sol :
$=\frac{\frac{1}{1}-\frac{1}{4}}{\frac{2}{1}-\frac{3}{4}}=\frac{3}{5}$
$=\frac{\frac{4-1}{4}}{\frac{8-3}{4}}$
$=\frac{3}{5}$
Question 35
यदि $\tan \theta=\frac{x}{y}$ , तब xsinθ+ycosθ का मान ज्ञात करें।
Sol :
tan θ=लम्ब/आधार$=\frac{x}{y}==\frac{AB}{BC}$
$A C=\sqrt{(AB)^{2}+(BC)^{2}}$
$A C=\sqrt{x^{2}+y^{2}}$
sin θ=लम्ब/कर्ण$=\frac{AB}{AC}=\frac{x}{\sqrt{x^{2}+y^{2}}}$
cos θ=आधार/कर्ण$=\frac{BC}{AC}=\frac{y}{\sqrt{x^{2}+y^{2}}}$
xsinθ+ycosθ
$x \times \frac{x}{\sqrt{x^{2}+y^{2}}}+y \times \frac{y}{\sqrt{x^{2}+y^{2}}}$
$\frac{x}{\sqrt{x^{2}+y^{2}}}+\frac{y}{\sqrt{x^{2}+y^{2}}}$
$=\frac{x^{2}+y^{2}}{\sqrt{x^{2}+y^{2}}}=\frac{\sqrt{x^{2}+y^{2}} \times \sqrt{x^{2}+y^{2}}}{\sqrt{x^{2}+y^{2}}}$
$=\sqrt{x^{2}+y^{2}}$
Question 36
यदि $\sin \theta=\frac{3}{5}$ , तब tan2θ+sinθcosθ+cotθ का मान ज्ञात करें।
Sol :
sin θ=लम्ब/कर्ण$=\frac{3}{5}=\frac{AB}{AC}$
$BC=\sqrt{(a c)^{2}-(a c)^{2}}$
$BC=\sqrt{(5)^{2}-(3)^{2}}$
$BC=\sqrt{25-9}=\sqrt{16}$
BC=4
tan θ=लम्ब/आधार$=\frac{AB}{BC}=\frac{3}{4}$
cos θ=आधार/कर्ण$=\frac{BC}{AC}=\frac{4}{5}$
cot θ=आधार/लम्ब$=\frac{BC}{AB}=\frac{4}{3}$
$\tan ^{2} \theta+\sin \theta \times \cos \theta+\cot \theta$
$=\left(\frac{3}{4}\right)^{2}+\frac{3}{5} \times \frac{4}{5}+\frac{4}{3}$
$=\frac{9}{16}+\frac{12}{25}+\frac{4}{3}$
$=\frac{2851}{1200}$
Question 37
यदि 4cotθ=3, तो दिखायें कि $\frac{\sin \theta+\cos \theta}{\sin \theta-\cos \theta}=7$
Sol :
Question 38
यदि $\sin \theta=\frac{m}{\sqrt{m^{2}+n^{2}}}$ , सिद्ध करें कि msinθ+ncosθ$=\sqrt{m^{2}+n^{2}}$
Sol :
$m \times \frac{m}{\sqrt{m^{2}+n^{2}}}+n \times \frac{n}{\sqrt{m^{2}+n^{2}}}$
$\frac{m^{2}}{\sqrt{m^{2}+n^{2}}}+\frac{n^{2}}{\sqrt{m^{2}+n^{2}}}$
$=\frac{m^{2}+n^{2}}{\sqrt{m^{2}+n^{2}}}=\frac{\sqrt{m^{2}+n^{2}} \times \sqrt{m^{2}+n^{2}}}{\sqrt{m^{2}+n^{2}}}$
$=\sqrt{m^{2}+n^{2}}$
LHS=RHS
Question 39
यदि $\cos \alpha=\frac{12}{13}$ , तो सिद्ध करें कि sin⍺(1-tan⍺)$=\frac{35}{156}$
Sol :
cos ⍺=आधार/कर्ण$=\frac{12}{13}=\frac{BC}{AC}$
$\mathrm{AB}=\sqrt{(a c)^{2}+(b c)^{2}}$
$\mathrm{AB}=\sqrt{(13)^{2}+(12)^{2}}$
$\mathrm{AB}=\sqrt{169+144}=\sqrt{25}$
AB=5
sin ⍺=लम्ब/कर्ण$=\frac{AB}{AC}=\frac{5}{13}$
tan ⍺=लम्ब/आधार$=\frac{AB}{BC}=\frac{5}{12}$
LHS
sin ⍺(1-tan ⍺)$=\frac{35}{156}$
$\frac{5}{13}\left(1-\frac{5}{12}\right)$
$=\frac{35}{156}=\frac{5}{13}\left(\frac{12-5}{12}\right)=\frac{35}{156}$
$=\frac{5}{13} \quad\left(\frac{17}{12}\right)=\frac{35}{156}$
$=\frac{35}{156}$
LHS=RHS
Question 40
यदि $q \cos \theta=\sqrt{q^{2}-p^{2}}$, तो सिद्ध करें कि $q \sin \theta=p$
Sol :
cos θ=आधार/कर्ण$=\frac{\sqrt{q^{2}}}{13}=\frac{BC}{AC}$
$\mathrm{AB}=\sqrt{(AC)^{2}-(BC)^{2}}$
$\mathrm{AB}=\sqrt{(q)^{2}-\left(\sqrt{q^{2}-p^{2}}\right)^{2}}$
$\mathrm{AB}=\sqrt{q^{2}-q^{2}+p^{2}}=\sqrt{p^{2}}$
AB=p
sin θ=लम्ब/कर्ण$=\frac{AB}{AC}=\frac{p}{q}$
LHS
q.sin θ=p
$q \times \frac{p}{q}=p$
p=RHS
LHS=RHS
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