If tan θ + sec θ = 3 \tan \theta + \sec \theta = \sqrt{3} tan θ + sec θ = 3 ( θ + π 6 ) \left(\theta + \dfrac{\pi}{6}\right) ( θ + 6 π )
হানি নাটস
Solution:
tan θ + sec θ = 3 \tan \theta+\sec \theta=\sqrt{3} tan θ + sec θ = 3
( tan θ + sec θ ) 2 = ( 3 ) 2 (\tan \theta+\sec \theta)^{2}=(\sqrt{3})^{2} ( tan θ + sec θ ) 2 = ( 3 ) 2
tan 2 θ + sec 2 θ + 2 tan θ sec θ = 3 \tan ^{2} \theta+\sec ^{2} \theta+2 \tan \theta \sec \theta=3 tan 2 θ + sec 2 θ + 2 tan θ sec θ = 3 [ sec 2 θ = 1 + tan 2 θ ] \left[\sec ^{2} \theta=1+\tan ^{2} \theta\right] [ sec 2 θ = 1 + tan 2 θ ]
tan 2 θ + ( 1 + tan 2 θ ) + 2 sin θ cos θ 1 cos θ = 3 \tan ^{2} \theta+\left(1+\tan ^{2} \theta\right)+2 \frac{\sin \theta}{\cos \theta} \frac{1}{\cos \theta}=3 tan 2 θ + ( 1 + tan 2 θ ) + 2 c o s θ s i n θ c o s θ 1 = 3
1 + 2 tan 2 θ + 2 sin θ cos 2 θ = 3 1+2 \tan ^{2} \theta+\frac{2 \sin \theta}{\cos ^{2} \theta}=3 1 + 2 tan 2 θ + c o s 2 θ 2 s i n θ = 3
2 tan 2 θ + 2 sin θ cos 2 θ = 2 2 \tan ^{2} \theta+\frac{2 \sin \theta}{\cos ^{2} \theta}=2 2 tan 2 θ + c o s 2 θ 2 s i n θ = 2
2 sin 2 θ cos 2 θ + 2 sin θ cos 2 θ = 2 2 \frac{\sin ^{2} \theta}{\cos ^{2} \theta}+\frac{2 \sin \theta}{\cos ^{2} \theta}=2 2 c o s 2 θ s i n 2 θ + c o s 2 θ 2 s i n θ = 2
sin 2 θ + sin θ cos 2 θ = 1 \frac{\sin ^{2} \theta+\sin \theta}{\cos ^{2} \theta}=1 c o s 2 θ s i n 2 θ + s i n θ = 1
sin 2 θ + sin θ = cos 2 θ sin θ = cos 2 θ − sin 2 θ sin θ = cos 2 θ sin θ = 1 − 2 sin 2 θ 2 sin 2 θ + sin θ − 1 = 0 \begin{aligned} \sin ^{2} \theta+\sin \theta & =\cos ^{2} \theta \\ \sin \theta & =\cos ^{2} \theta-\sin ^{2} \theta \\ \sin \theta & =\cos ^{2} \theta \\ \sin \theta & =1-2 \sin ^{2} \theta \\ 2 \sin ^{2} \theta & +\sin \theta-1=0 \end{aligned} sin 2 θ + sin θ sin θ sin θ sin θ 2 sin 2 θ = cos 2 θ = cos 2 θ − sin 2 θ = cos 2 θ = 1 − 2 sin 2 θ + sin θ − 1 = 0
2 sin 2 θ + 2 sin θ − sin θ − 1 = 0 2 sin θ ( sin θ + 1 ) − 1 ( sin θ + 1 ) = 0 ( 2 sin θ − 1 ) ( sin θ + 1 ) = 0 2 sin θ − 1 = 0 ar sin θ + 1 = 0 2 sin θ = 1 sin θ = − 1 sin θ = 1 2 principal ualue is π 6 + π 6 = π 3 θ = sin − 1 ( 1 2 ) \begin{array}{l} 2 \sin ^{2} \theta+2 \sin \theta-\sin \theta-1=0 \\ 2 \sin \theta(\sin \theta+1)-1(\sin \theta+1)=0 \\ (2 \sin \theta-1)(\sin \theta+1)=0 \\ 2 \sin \theta-1=0 \text { ar } \sin \theta+1=0 \\ 2 \sin \theta=1 \quad \sin \theta=-1 \\ \sin \theta=\frac{1}{2} \quad \text { principal ualue is } \frac{\pi}{6}+\frac{\pi}{6}=\frac{\pi}{3} \\ \theta=\sin ^{-1}\left(\frac{1}{2}\right) \end{array} 2 sin 2 θ + 2 sin θ − sin θ − 1 = 0 2 sin θ ( sin θ + 1 ) − 1 ( sin θ + 1 ) = 0 ( 2 sin θ − 1 ) ( sin θ + 1 ) = 0 2 sin θ − 1 = 0 ar sin θ + 1 = 0 2 sin θ = 1 sin θ = − 1 sin θ = 2 1 principal ualue is 6 π + 6 π = 3 π θ = sin − 1 ( 2 1 )
