KC Sinha Mathematics Solution Class 10 Chapter 8 त्रिकोणमितीय अनुपात एवम सर्वसमिकाए ( Trigonometry Ratios and Identities ) Exercise 8.4 (Q31-Q40)

Exercise 8.4

Question 31

$\tan ^{2} \phi-\sin ^{2} \phi-\tan ^{2} \phi \cdot \sin ^{2} \phi=0$

Sol :

LHS

$\tan ^{2} \phi-\sin ^{2} \phi-\tan ^{2} \phi \times \sin ^{2} \phi$

$\frac{\sin ^{2} \phi}{\cos ^{2} \phi}-\sin ^{2} \phi-\frac{\sin ^{2} \phi}{\cos ^{2} \phi} \times \sin ^{2} \phi$

$\sin ^{2} \phi\left[\frac{1}{\cos ^{2} \phi}-1\right]-\frac{\sin ^{4} \phi}{\cos ^{2} \phi}$

$\sin ^{2} \phi\left[\frac{1-\cos ^{2} \phi}{\cos ^{2} \phi}\right]-\frac{\sin ^{4} \phi}{\cos ^{2} \phi}$

$\sin ^{2} \phi\left[\frac{\sin ^{2} \phi}{\cos ^{2} \phi}\right]-\frac{\sin ^{4} \phi}{\cos ^{2} \phi}$

$\frac{\sin ^{4} \phi}{\cos ^{2} \phi}-\frac{\sin ^{4} \phi}{\cos ^{2} \phi}$

=0 proved

Question 32

$\tan ^{2} \phi+\cot ^{2} \phi+2=\sec ^{2} \phi \cdot \operatorname{cosec}^{2} \phi$

Sol :

LHS

$\tan ^{2} \phi+\cot ^{2} \phi+2$

$\sin ^{2} \phi-1+\operatorname{cosec}^{2} \phi-1+2$

$\frac{1}{\cos ^{2} \phi}-1+\frac{1}{\sin ^{2} \phi}-1+2$

$\frac{1}{\cos ^{2} \phi}+\frac{1}{\sin ^{2} \phi}-\not 2+\not 2$

$\frac{\sin ^{2} \phi+\cos ^{2} \phi}{\cos ^{2} \phi \cdot \sin ^{2} \phi}=\frac{1}{\cos ^{2} \phi \cdot \sin ^{2} \phi}$

$\sin ^{2} \phi \times \operatorname{cosec}^{2} \phi$ proved

Question 33

$\frac{\operatorname{cosec} \theta+\cot \theta-1}{\cot \theta-\operatorname{cosec} \theta+1}=\frac{1+\cos \theta}{\sin \theta}$

Sol :

$\frac{\operatorname{cosec} \theta+\cot \theta-1}{\cot \theta-\operatorname{cosec} \theta+1}$

$\frac{\operatorname{cosec} \theta+\cot \theta-\left(\operatorname{cosec}^{2} \theta-\cot ^{2} \theta\right)}{\cot \theta-\operatorname{cosec} \theta+1}$

$\frac{(\operatorname{cosec} \theta+\cot \theta)-(\operatorname{cosec} \theta-\cot \theta) \times(\operatorname{cosec} \theta+\cot \theta)}{1-\operatorname{cosec} \theta+\cot \theta}$

$\frac{(\operatorname{cosec} \theta+\cot \theta) \times[1-(\operatorname{cosec} \theta-\cot \theta)]}{1-\operatorname{cosec} \theta+\cot \theta}$

$\frac{(\operatorname{cosec} \theta+\cot \theta) \times(1-\operatorname{cosec} \theta+\cot \theta)}{1-\operatorname{cosec} \theta+\cot \theta}$

$=\operatorname{cosec} \theta+\cot \theta$

$=\frac{1}{\sin \theta}+\frac{\cos \theta}{\sin \theta}$

$=\frac{1+\cos \theta}{\sin \theta}=$ proved

Question 34

$\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}=1+\tan \theta+\cot \theta$

Sol :

LHS

$\frac{\tan \theta}{1-\cot \theta}+\frac{\cot \theta}{1-\tan \theta}$

$\frac{\frac{\sin \theta}{\cos \theta}}{1-\frac{\cos \theta}{\sin \theta}}+\frac{\frac{\cos \theta}{\sin \theta}}{1-\frac{\sin \theta}{\cos \theta}}$

$\frac{\frac{\sin \theta}{\cos \theta}}{\frac{\sin \theta-\cos \theta}{\sin \theta}}+\frac{\frac{\cos \theta}{\sin \theta}}{\frac{\cos \theta-\sin \theta}{\cos \theta}}$

$\frac{\sin ^{2} \theta}{\cos \theta(\sin \theta-\cos \theta)}+\frac{\cos ^{2} \theta}{\sin \theta(\cos \theta-\sin \theta)}$

$\frac{\sin ^{2} \theta}{\cos \theta(\sin \theta-\cos \theta)}-\frac{\cos ^{2} \theta}{\sin \theta(\sin \theta-\cos \theta)}$

$\frac{\sin ^{2} \theta-\cos ^{2} \theta}{\cos \theta \sin \theta(\sin \theta-\cos \theta)}$

$\frac{(\sin \theta-\cos \theta) \times\left(\sin ^{2} \theta+\cos ^{2} \theta+\sin \theta \times \cos \theta\right)}{\cos \theta \times \cot \theta(\sin \theta-\cos \theta)}$

$\frac{\sin \theta \times \cos \theta+\sin ^{2} \theta+\cos ^{2} \theta}{\cos \theta \times \sin \theta}$

$\frac{\sin \theta \times \cos \theta}{\cos \theta \times \sin \theta}+\frac{\sin ^{2} \theta}{\cos \theta \times \sin \theta}+\frac{\cos ^{2} \theta}{\cos \theta \times \sin \theta}$

$1+\tan \theta+\cot \theta$ proved

Question 35

$\frac{1-\cos \theta}{1+\cos \theta}=(\cot \theta-\operatorname{cosec} \theta)^{2}$

Sol :

LHS

$\frac{1-\cos \theta}{1+\cos \theta}$

परिमेयकरण करने पर

$\frac{1-\cos \theta}{1+\cos \theta} \times \frac{1-\cos \theta}{1-\cos \theta}$

$\frac{(1-\cos \theta)^{2}}{1^{2}-\cos ^{2} \theta}$

$\frac{(1-\cos \theta)^{2}}{1-\cos ^{2} \theta}=\frac{(1-\cos \theta)^{2}}{\sin ^{2} \theta}$

$\left(\frac{1-\cos \theta}{\sin \theta}\right)^{2}$

$\left(\frac{1-\cos \theta}{\sin \theta}-\frac{\cos \theta}{\sin \theta}\right)^{2}$

$=(\cos \theta-\operatorname{cosec} \theta)^{2}$

निम्नलिखित सर्वसमिकाओं को सिद्ध करें :

Question 36

$\sqrt{\frac{1-\cos \theta}{1+\cos \theta}}=\frac{1-\cos \theta}{\sin \theta}$

Sol :

LHS

$\sqrt{\frac{1+\cos \theta}{1-\cos \theta}}$

परिमेयकरण करने पर

$\sqrt{\frac{1+\cos \theta}{1-\cos \theta} \times \frac{1+\cos \theta}{1+\cos \theta}}$

$\sqrt{\frac{(1+\cos \theta)^{2}}{1^{2}-\cos ^{2} \theta}}$

$\sqrt{\frac{(1+\cos \theta)^{2}}{1-\cos ^{2} \theta}}$

$\sqrt{\frac{(1+\cos \theta)^{2}}{\sin ^{2} \theta}}$

$\sqrt{\left(\frac{1+\cos \theta}{\sin \theta}\right)^{2}}$

$\frac{1+\cos \theta}{\sin \theta}$=Proved

Question 37

$\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}=\frac{\cos \theta}{1+\sin \theta}$

Sol :

LHS

$\sqrt{\frac{1-\cos \theta}{1+\cos \theta}}$

परिमेयकरण करने पर

$\sqrt{\frac{1-\cos \theta}{1+\cos \theta} \times \frac{1-\cos \theta}{1-\cos \theta}}$

$\sqrt{\frac{(1-\cos \theta)^{2}}{1^{2}-\cos ^{2} \theta}}$

$\sqrt{\frac{(1-\cos \theta)^{2}}{1-\cos ^{2} \theta}}$

$\sqrt{\frac{(1-\cos \theta)^{2}}{\sin ^{2} \theta}}$

$\sqrt{\left(\frac{1-\cos \theta}{\sin \theta}\right)^{2}}$

$\frac{1-\cos \theta}{\sin \theta}=$ proved

Question 38

$\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}=\sec \theta-\tan \theta$

Sol :

LHS

$\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}$

परिमेयकरण करने पर

$\sqrt{\frac{1-\sin \theta}{1+\sin \theta}} \times \frac{1-\sin \theta}{1-\sin \theta}$

$\sqrt{\frac{(1-\sin \theta)^{2}}{1^{2}+\sin ^{2} \theta}}=\sqrt{\frac{(1-\sin \theta)^{2}}{1+\sin ^{2} \theta}}$

$\sqrt{\frac{(1-\sin \theta)^{2}}{\cos ^{2} \theta}}=\sqrt{\left(\frac{1-\sin \theta}{\cos \theta}\right)^{2}}$

$\frac{1-\sin \theta}{\cos \theta}=\frac{1}{\cos \theta}-\frac{\sin \theta}{\cos \theta}$

$\sec \theta-\tan \theta$=RHS

Question 39

$\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}=\frac{\cos \theta}{1+\sin \theta}$

Sol :

LHS

$\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}$

अंश का परिमेयकरण करने पर

$\sqrt{\frac{1-\sin \theta}{1+\sin \theta} \times \frac{1+\sin \theta}{1+\sin \theta}}$

$\sqrt{\frac{1^{2}-\sin ^{2} \theta}{(1+\sin \theta)^{2}}}=\sqrt{\frac{1-\sin ^{2} \theta}{(1+\sin \theta)^{2}}}$

$\sqrt{\frac{\cos ^{2} \theta}{(1+\sin \theta)^{2}}}=\sqrt{\left(\frac{\cos \theta}{1+\sin \theta}\right)^{2}}$

$\frac{\cos \theta}{1+\sin \theta}=\text { R.H.S. }$

Question 40

$\sqrt{\frac{1+\sin \theta}{1-\sin \theta}}+\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}=2 \sec \theta$

Sol :

LHS

$\sqrt{\frac{1+\sin \theta}{1-\sin \theta}}+\sqrt{\frac{1-\sin \theta}{1+\sin \theta}}$

$\frac{\sqrt{1+\sin \theta}}{\sqrt{1-\sin \theta}}+\frac{\sqrt{1-\sin \theta}}{\sqrt{1+\sin \theta}}$

$\frac{(\sqrt{1+\sin \theta})^{2}+(\sqrt{1-\sin \theta})^{2}}{(\sqrt{1-\sin \theta})(\sqrt{1+\sin \theta})}$

$\frac{(1+\sin \theta)+(1-\sin \theta)}{(\sqrt{1-\sin \theta})(\sqrt{1+\sin \theta})}$

$\frac{1+\sin \theta+1-\sin \theta}{\sqrt{1^{2}-\sin ^{2} \theta}}$

$\frac{2}{\sqrt{1-\sin ^{2} \theta}}=\frac{2}{\cos \theta}$

$2 \times \frac{1}{\cos \theta}=2 \sec \theta$=proved