KC Sinha Mathematics Solution Class 10 Chapter 8 त्रिकोणमितीय अनुपात एवम सर्वसमिकाए ( Trigonometry Ratios and Identities ) Exercise 8.1 (Q31-40)

 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 :

tan ⍺=लम्ब/आधार$=\frac{4}{5}=\frac{AB}{BC}$

$\mathrm{AC}=\sqrt{(a b)^{2}+(b c)^{2}}$

$\mathrm{AC}=\sqrt{(4)^{2}+(5)^{2}}=\sqrt{16+25}$

$\mathrm{AC}=\sqrt{41}$

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 :

cot θ=आधार/लम्ब$=\frac{1}{\sqrt{3}}=\frac{BC}{AB}$

tan ⍺=लम्ब/आधार$=\frac{4}{5}=\frac{AB}{BC}$

$\mathrm{AC}=\sqrt{(a b)^{2}+(b c)^{2}}$

$\mathrm{AC}=\sqrt{(4)^{2}+(5)^{2}}=\sqrt{16+25}$

$\mathrm{AC}=\sqrt{41}$

sin ⍺=लम्ब/कर्ण$=\frac{AB}{AC}=\frac{4}{\sqrt{41}}$

cos ⍺=आधार/कर्ण$=\frac{BC}{AC}=\frac{5}{\sqrt{41}}$

LHS

$\frac{1-\operatorname{cos}^{2} \theta}{2-\sin ^{2} \theta}=\frac{3}{5}$

$=\frac{1-\left(\frac{1}{2}\right)^{2}}{2-\left(\frac{\sqrt{3}}{2}\right)^{2}}=\frac{3}{5}$

$=\frac{\frac{1}{1}-\frac{1}{4}}{\frac{2}{1}-\frac{3}{4}}=\frac{3}{5}$

$=\frac{\frac{4-1}{4}}{\frac{8-3}{4}}$

$=\frac{3}{5}$ 

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 :

cot θ=आधार/लम्ब$=\frac{3}{4}=\frac{BC}{AB}$

$\mathrm{AC}=\sqrt{(a b)^{2}+(b c)^{2}}$

$\mathrm{AC}=\sqrt{(4)^{2}+(3)^{2}}$

$\mathrm{AC}=\sqrt{16+9}=\sqrt{25}$

AC=5

sin θ=लम्ब/कर्ण$=\frac{AB}{AC}=\frac{4}{5}$

cos θ=आधार/कर्ण$=\frac{BC}{AC}=\frac{3}{5}$

LHS

$\frac{\sin \theta+\cos \theta}{\sin \theta+\cos \theta}=\frac{\frac{4}{5}+\frac{3}{5}}{\frac{4}{5}-\frac{3}{5}}$

$=\frac{\frac{4+3}{5}}{\frac{4-3}{5}}=\frac{\frac{7}{5}}{\frac{1}{5}}$

=7

Question 38

यदि $\sin \theta=\frac{m}{\sqrt{m^{2}+n^{2}}}$ , सिद्ध करें कि msinθ+ncosθ$=\sqrt{m^{2}+n^{2}}$

Sol :

sin θ=लम्ब/कर्ण$=\frac{m}{\sqrt{m^{2}+n^{2}}}=\frac{a b}{a c}$

$\mathrm{BC}=\sqrt{(AC)^{2}-(AB)^{2}}$

$\mathrm{BC}=\sqrt{\left(\sqrt{m^{2}+n^{2}}\right)^{2}-m^{2}}$

$\mathrm{BC}=\sqrt{m^{2}+n^{2}-m^{2}}=\sqrt{n^{2}}$

BC=n

cos θ=आधार/कर्ण$=\frac{BC }{AC}=\frac{n}{\sqrt{m^{2}+n^{2}}}$

LHS

m.sin+θ n.cos θ$=\sqrt{m^{2}+n^{2}}$

$m \times \frac{m}{\sqrt{m^{2}+n^{2}}}+n \times \frac{n}{\sqrt{m^{2}+n^{2}}}$

$\frac{m^{2}}{\sqrt{m^{2}+n^{2}}}+\frac{n^{2}}{\sqrt{m^{2}+n^{2}}}$

$=\frac{m^{2}+n^{2}}{\sqrt{m^{2}+n^{2}}}=\frac{\sqrt{m^{2}+n^{2}} \times \sqrt{m^{2}+n^{2}}}{\sqrt{m^{2}+n^{2}}}$

$=\sqrt{m^{2}+n^{2}}$

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