Exercise 8.4
Type IV : शंर्त वाले सर्वसमिकाओं पर आधारित प्रश्न :
यदि \sec \theta+\tan \theta=m और \sec \theta-\tan \theta=n, तो सिद्ध
करें कि \sqrt{m n}=1
Sol :
LHS
\sqrt{m n}=\sqrt{\sec \theta+\tan \theta(\sec \theta-\tan \theta)}
\sqrt{\sec ^{2} \theta-\sec \theta \times \tan \theta+\sec \theta \times \tan \theta-\tan ^{2} \theta}
\sqrt{\sec ^{2} \theta-\tan ^{2} \theta}=\sqrt{1}
=1 proved
यदि \cos \theta+\sin \theta=1, तब सिद्ध करें कि \cos \theta-\sin
\theta=\pm 1
Sol :
माना -\cos \theta-\sin \theta=\mathrm{x}
दिया है - \cos \theta+\sin \theta=1
दोनों को वर्ग करने पर
(\cos \theta-\sin \theta)^{2}+(\cos \theta+\sin \theta)^{2}=x^{2}+1^{2}
\cos ^{2} \theta+\sin ^{2} \theta-2 \cos \theta \cdot \sin \theta+\cos ^{2} \theta+\sin ^{2} \theta+2 \cos \theta \cdot \sin \theta=x^{2}+1^{2}
\cos ^{2} \theta+\sin ^{2} \theta+\cos ^{2} \theta+\sin ^{2} \theta=x^{2}+1
1+1=x^{2}+1
2=x^{2}+1
x^{2}+1=2
x^{2}=2-1
x=\sqrt{1}
x=\pm 1
\cos \theta+\sin \theta=x
\cos \theta+\sin \theta=\pm 1
यदि \sin \theta+\sin ^{2} \theta=1. तब सिद्ध करें कि \cos ^{2}
\theta+\cos ^{4} \theta=1
Sol :
\sin \theta+\sin ^{2} \theta=1
\sin \theta=1-\sin ^{2} \theta
\sin \theta+\cos ^{2} \theta
LHS
\cos ^{2} \theta+\cos ^{4} \theta
\cos ^{2} \theta+\left(\cos ^{2} \theta\right)^{2}
\cos ^{2} \theta+\sin ^{2} \theta
=1 proved
यदि \tan \theta+\sec \theta=x. सिद्ध करें कि \sin
\theta=\frac{x^{2}-1}{x^{2}+1}
Sol :
\sec \theta+\tan \theta=\mathrm{x}....(i)
हम जानते है कि \sec ^{2} \theta-\tan ^{2} \theta=1
(\sec \theta-\tan \theta)(\sec \theta+\tan \theta)=1
(\sec \theta-\tan \theta) \cdot x=1
\sec \theta-\tan \theta=\frac{1}{x}...(ii)
सामीकरण (i) तथा (ii) से
\sec \theta+\tan \theta=x
\sec \theta-\tan \theta=\frac{1}{x}
2 \sec \theta=x+\frac{1}{x}
2 \sec \theta=\frac{x^{2}+1}{x}
\sec \theta=\frac{x^{2}+1}{2 x}=\frac{\mathrm{AC}}{\mathrm{BC}}
\mathrm{AB}=\sqrt{(\mathrm{AC})^{2}-(\mathrm{BC})^{2}}
\mathrm{AB}=\sqrt{\left(x^{2}+1\right)^{2}-(2 \mathrm{x})^{2}}
\mathrm{AB}=\sqrt{\left(\left(x^{2}\right)^{2}+(1)^{2}+2 \cdot x^{2} \cdot 1\right)-4 x^{2}}
A B=\sqrt{\left(x^{2}\right)^{2}+1+2 x^{2}-4 x^{2}}
A B=\sqrt{\left(x^{2}\right)^{2}-2 x^{2}+1}
A B=\sqrt{\left(x^{2}-1\right)^{2}}
A B=x^{2}-1
\sin \theta=\frac{A B}{A C}=\frac{x^{2}-1}{x^{2}+1}
\sin \theta=\frac{x^{2}-1}{x^{2}+1}
यदि \sin \theta+\cos \theta=p और \sec \theta+\operatorname{cosec}
\theta=q, तब सिद्ध करें कि q\left(p^{2}-1\right)=2 p
Sol :
LHS
\mathrm{q}\left(\mathrm{p}^{2}-1\right)
(\sec \theta+\operatorname{cosec} \theta)\left[(\sin \theta+\cos \theta)^{2}-1\right]
\frac{1}{\cos \theta}+\frac{1}{\sin \theta}\left[\sin ^{2} \theta+\cos \theta^{2}+2 \sin \theta \cdot \cos \theta-1\right]
\frac{\sin \theta+\cos \theta}{\cos \theta \cdot \sin \theta}[1+2 \sin \theta \cdot \cos \theta- 1]
\frac{\sin \theta+\cos \theta}{\cos \theta \cdot \sin \theta} \times 2 \sin \theta \cdot \cos \theta
=2p proved
यदि x \cos \theta=a और y=a \tan \theta, तब सिद्ध करें कि
x^{2}-y^{2}=a^{2}
Sol :
x \cos \theta=a
x=\frac{a}{\cos \theta}
x=a \sec \theta, \quad y=a \tan \theta
दोनों को वर्ग करके घटाने पर
x^{2}-y^{2}=(a \sec \theta)^{2}-(a \tan \theta)^{2}
x^{2}-y^{2}=a^{2} \sec ^{2} \theta-a^{2} \tan ^{2} \theta
x^{2}-y^{2}=a^{2}\left(\sec ^{2} \theta-\tan ^{2} \theta\right)
x^{2}-y^{2}=a^{2}(1)^{2}
x^{2}-y^{2}=a^{2}
Question 47 (to be fixed)
यदि x=r \cos \alpha. \sin \beta, y=r \cos \alpha \cdot \cos \beta, z=r
\sin \alpha, तब सिद्ध करें कि x^{2}+y^{2}+z^{2}=r ?
Sol :
LHS
x^{2}+y^{2}+z^{2}=(r \cos \alpha \cdot \sin \beta)^{2}+(r \cos \alpha \cdot \cos \beta)^{2}+(r \sin \alpha)^{2}
x^{2}+y^{2}+z^{2}=r^{2} \cos ^{2} \alpha \cdot \sin ^{2} \beta+r^{2} \cos ^{2} \alpha \cdot \cos ^{2} \beta+r \sin ^{2} \alpha
x^{2}+y^{2}+z^{2}=r^{2} \cos ^{2} \alpha\left(\sin ^{2} \beta+\cos ^{2} \beta\right)+r^{2} \sin ^{2} \alpha
x^{2}+y^{2}+z^{2}=r^{2} \cos ^{2} \alpha(1)+r^{2} \cdot \sin ^{2} \alpha
x^{2}+y^{2}+z^{2}=r^{2} \cos ^{2} \alpha+r^{2} \cdot \sin ^{2} \alpha
x^{2}+y^{2}+z^{2}=r^{2}\left(\cos ^{2} \alpha+\sin ^{2} \alpha\right)
x^{2}+y^{2}+z^{2}=r^{2}(1)^{2}
x^{2}+y^{2}+z^{2}=r^{2}= R.H.S.
यदि \sec \theta-\tan \theta=x, तब सिद्ध करें कि
(i) \cos \theta=\frac{2 x}{1+x^{2}}
(ii) \sin \theta=\frac{1-x^{2}}{1+x^{2}}
Sol :
\sec \theta-\tan \theta=\mathrm{x}...(i)
हम जानते है कि \left(\sec ^{2} \theta-\tan ^{2} \theta\right)=1
(\sec \theta+\tan \theta)(\sec \theta-\tan \theta)=1
(\sec \theta+\tan \theta) \cdot \mathrm{x}=1
\sec \theta+\tan \theta=\frac{1}{x}- (ii)
सामीकरण (i) तथा (ii) से
\sec \theta-\tan \theta=x
\sec \theta+\tan \theta=\frac{1}{x}
2 \sec \theta=x+\frac{1}{x}
2 \sec \theta=\frac{x^{2}+1}{x}
\sec \theta=\frac{x^{2}+1}{2 x}=\frac{\mathrm{AC}}{\mathrm{BC}}
A B=\sqrt{(A C)^{2}-(B C)^{2}}
A B=\sqrt{\left(x^{2}+1\right)^{2}-(2 x)^{2}}
A B=\sqrt{\left(x^{2}\right)^{2}+(1)^{2}+2 x^{2} \cdot 1-4 x^{2}}
A B=\sqrt{\left(x^{2}\right)^{2}+1+2 x^{2}-4 x^{2}}
A B=\sqrt{\left(x^{2}\right)^{2}+1-2 x^{2}}
A B=\sqrt{\left(x^{2}-1\right)^{2}}
A B=x^{2}-1
\cos \theta=\frac{\mathrm{BC}}{\mathrm{AC}}=\frac{2 x}{x^{2}+1}
\sin \theta=\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{x^{2}-1}{x^{2}+1}
यदि a \cos \theta+b \sin \theta=c, तब सिद्ध करें कि a \sin \theta-b
\cos \theta=\pm \sqrt{a^{2}+b^{2}-c^{2}}
Sol :
माना -\mathrm{a} \sin \theta-\mathrm{b} \cos \theta=\mathrm{x}...(i)
दिया है - \mathrm{a} \cos \theta+\mathrm{b} \sin \theta=\mathrm{c}...(ii)
दोनों को वर्ग करके जोड़ने पर
(a \sin \theta-b \cos \theta)^{2}+(a \cos \theta+b \sin \theta)^{2}=x^{2}+c^{2}
a^{2} \sin ^{2} \theta+b^{2} \cos ^{2} \theta-2 a \sin \theta \cdot b \cos \theta+a^{2} \cos ^{2} \theta+b^{2} \sin ^{2} \theta+ 2 a \cos \theta \cdot b \sin \theta=x^{2}+c^{2}
a^{2} \sin ^{2} \theta+b^{2} \cos ^{2} \theta+a^{2} \cos ^{2} \theta+b^{2} \sin ^{2} \theta=x^{2}+c^{2}
\left(a^{2} \sin ^{2} \theta+a^{2} \cos ^{2} \theta\right)+\left(b^{2} \cos ^{2} \theta+b^{2} \sin ^{2} \theta\right)=x^{2}+c^{2}
a^{2}\left(\sin ^{2} \theta+\cos ^{2} \theta\right)+b^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)=x^{2}+c^{2}
a^{2}=1+b^{2} \times 1=x^{2}+c^{2}
a^{2}+b^{2}=x^{2}+c^{2}
x^{2}+c^{2}=a^{2}+b^{2}
x^{2}=a^{2}+b^{2}-c^{2}
x=\pm \sqrt{a^{2}+b^{2}-c^{2}}
x=\pm \sqrt{a^{2}+b^{2}-c^{2}}
यदि 1+\sin ^{2} \theta=3 \sin \theta \cdot \cos \theta, तब सिद्ध करें
कि \tan \theta=1 या, \frac{1}{2}, जहाँ \theta<90^{\circ}
Sol :
1+\sin ^{2} \theta-3 \sin \theta \cos \theta
दोनों तरफ \cos ^{2} \theta से भाग देने पर
\frac{1}{\cos ^{2} \theta}+\frac{\sin ^{2} \theta}{\cos ^{2} \theta}=\frac{3 \cdot \sin \theta \cos \theta}{\cos ^{2} \theta}
\sec ^{2} \theta+\tan ^{2} \theta=3 \tan \theta
1+\tan ^{2} \theta+\tan ^{2} \theta-3 \tan \theta=0
1+2 \tan ^{2} \theta-3 \tan \theta=0
2 \tan ^{2} \theta-3 \tan \theta+1=0
2 \tan ^{2} \theta-2 \tan \theta-\tan \theta+1=0
2 \tan \theta(\tan \theta-1)-1(\tan \theta-1)=0
\begin{array}{r|rl}(\tan \theta-1) & 2 \tan \theta & -1 \\\tan \theta=1 & 2 \tan \theta & =1 \\&\tan \theta &=\frac{1}{2}\end{array}
\tan \theta=1 या \frac{1}{2}
यदि a \cos \theta-b \sin \theta=x और a \sin \theta+b \cos \theta=y,
तो सिद्ध करें कि a^{2}+b^{2}=x^{2}+y^{2}
Sol :
a \cos \theta-b \sin \theta=x....(i)
a \sin \theta+b \cos \theta=y.....(ii)
दोनों समीकरण को वर्ग करके जोड़ने पर
(a \cos \theta-b \sin \theta)^{2}+(a \sin \theta+b \cos \theta)=x^{2}+y^{2}
a^{2} \cos ^{2} \theta+b^{2} \sin ^{2} \theta-2 a \cos \theta \cdot b \sin \theta+a^{2} \sin ^{2} \theta+b^{2} \cos ^{2} \theta+2 a \sin \theta \cdot b \cos \theta=x^{2}+y^{2}
a^{2} \cos ^{2} \theta+b^{2} \sin ^{2} \theta+a^{2} \sin ^{2} \theta+b^{2} \cos ^{2} \theta=x^{2}+y^{2}
\left(a^{2} \cos ^{2} \theta+a^{2} \sin ^{2} \theta\right)+\left(b^{2} \sin ^{2} \theta+b^{2} \cos ^{2} \theta\right)=x^{2}+y^{2}
a^{2} \times 1+b^{2} \times 1=x^{2}+y^{2}
a^{2}+b^{2}=x^{2}+y^{2}
यदि x=a \sec \theta+b \tan \theta और y=a \tan \theta+b \sec \theta,
तो सिद्ध करें कि x^{2}-y^{2}=a^{2}-b^{2}
Sol :
x=a \sec \theta+b \tan \theta...(i)
y=a \tan \theta+b \sec \theta...(ii)
समीकरण (i) तथा (ii) को वर्ग करके घटाने पर
x^{2}-y^{2}=(a \sec \theta-b \tan \theta)^{2}-(a \tan \theta+b \sec \theta)^{2}
x^{2}-y^{2}=a^{2} \sec ^{2} \theta+b^{2} \tan ^{2} \theta+2 a \sec \theta \cdot b \tan \theta-\left(a^{2} \tan ^{2} \theta+b^{2} \sec ^{2} \theta+\right. 2 \operatorname{atan} \theta \cdot b \sec \theta)
x^{2}-y^{2}=a^{2} \sec ^{2} \theta+b^{2} \tan ^{2} \theta+2 a \sec \theta \cdot b \tan \theta-a^{2} \tan ^{2} \theta-b^{2} \sec ^{2} \theta-2 a \tan \theta \cdot b \sec \theta
x^{2}-y^{2}=a^{2} \sec ^{2} \theta+b^{2} \tan ^{2} \theta-a^{2} \tan ^{2} \theta-b^{2} \sec ^{2} \theta
x^{2}-y^{2}=a^{2} \sec ^{2} \theta-a^{2} \tan ^{2} \theta+b^{2} \tan ^{2} \theta-b^{2} \sec ^{2} \theta
x^{2}-y^{2}=a^{2}\left(\sec ^{2} \theta-\tan ^{2} \theta\right)-b^{2}\left(\tan ^{2} \theta-\sec ^{2} \theta\right)
x^{2}-y^{2}=a^{2}\left(\sec ^{2} \theta-\tan ^{2} \theta\right)-b^{2}\left(\sec ^{2} \theta-\tan ^{2} \theta\right)
x^{2}-y^{2}=a^{2}-b^{2}
यदि \left(a^{2}-b^{2}\right) \sin \theta+2 a b \cos \theta=a^{2}+b^{2},
तो सिद्ध करें कि \tan \theta=\frac{a^{2}-b^{2}}{2 a b}
Sol :
\left(a^{2}-b^{2}\right) \sin \theta+2 a b \cos \theta=a^{2}+b^{2}
दोनों तरफ \cos \theta से भाग देने पर
\left(a^{2}-b^{2}\right) \times \frac{\sin \theta}{\cos \theta}+(2 a b) \times \frac{\cos \theta}{\cos \theta}=\frac{a^{2}}{\cos \theta}
a^{2}-b^{2} \tan \theta+2 a b=a^{2} \sec \theta+b^{2} \sec \theta
a^{2}-b^{2} \tan \theta+2 \mathrm{ab}=a^{2}+b^{2}(\sec \theta)
a^{2}-b^{2} \tan \theta+2 \mathrm{ab}=\left(a^{2}+b^{2}\right) \sqrt{1+\tan ^{2} \theta}
दोनों तरफ वर्ग करने पर
\left[\left(a^{2}-b^{2}\right) \tan \theta+2 a b\right]^{2}=\left[\left(a^{2}-b^{2}\right) \sqrt{1+\tan ^{2}} \theta\right]^{2}
\left(a^{2}-b^{2} \tan \theta\right)^{2}+(2 a b)^{2}+2\left(a^{2}-b^{2}\right) \tan \theta \cdot 2 a b=\left(a^{2}+b^{2}\right)^{2}+\left(1+\tan ^{2} \theta\right)
\left(a^{2}-b^{2}\right) \times \tan ^{2} \theta+4 a b\left(a^{2}-b^{2}\right) \times \tan \theta+4 a^{2} b^{2}=\left(a^{2}+b^{2}\right)^{2}+\left(a^{2}+b^{2}\right)^{2} \times \tan ^{2} \theta
\left(a^{2}-b^{2}\right)^{2} \times \tan ^{2} \theta-\left(a^{2}-b^{2}\right)^{2} \tan ^{2} \theta+4 a b\left(a^{2}-b^{2}\right) \tan \theta+4 a^{2} b^{2}-\left(a^{2}+b^{2}\right)^{2}=0
\left[a^{4}-2 a^{2} \cdot b^{2}+b^{4}-a^{4}-2 a^{2} \cdot b^{2}-b^{4}\right] \tan ^{2} \theta+4 a b\left(a^{2}-b^{2}\right)\tan \theta+4 a^{2} \cdot b^{2}-a^{4}-b^{4} 2 a^{2} b^{2}=0
-4 a^{2} \cdot b^{2} \times \tan ^{2} \theta+4 a b\left(a^{2}-b^{2}\right) \tan \theta-a^{2}-b^{4}+2 a^{2} \cdot b^{2}=0
4 a^{2} \cdot b^{2} \times \tan ^{2} \theta-4 a b\left(a^{2}-b^{2}\right) \tan \theta+a^{4}+b^{4}-2 a^{2} \cdot b^{2}=0
4 a^{2} \cdot b^{2} \times \tan ^{2} \theta-4 a b\left(a^{2}-b^{2}\right) \tan\theta+\left(a^{2}\right)^{2}+\left(b^{2}\right)^{2}-2 a^{2} \cdot b^{2}=0
(2 a b \tan \theta)^{2}-2 \cdot 2 a b \tan \theta\left(a^{2}-b^{2}\right)+\left(a^{2}-b^{2}\right)^{2}=0
\left[2 a b \tan \theta-\left(a^{2}-b^{2}\right)\right]^{2}=0
2 a b \tan \theta-\left(a^{2}-b^{2}\right)=0
2 a b \tan \theta=a^{2}-b^{2}
\tan \theta=\frac{a^{2}-b^{2}}{2 a b}
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