सर्वसमिका sec ^{2} theta=1+tan ^{2} theta का | vedclass

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Question

Since we will apply the identity involving $\sec 	heta$ and $	an 	heta,$ let us first convert the $LHS$ (of the identity we need to prove) in terms

Vedclass · Accepted Answer

Vedclass · Answer

Since we will apply the identity involving $\sec 	heta$ and $	an 	heta,$ let us first convert the $LHS$ (of the identity we need to prove) in terms of $\sec 	heta$ and $	an 	heta$ by dividing numerator and denominator by $\cos 	heta .$

$LHS=\frac{\sin 	heta-\cos 	heta+1}{\sin 	heta+\cos 	heta-1}=\frac{	an 	heta-1+\sec 	heta}{	an 	heta+1-\sec 	heta}$

$=\frac{(	an 	heta+\sec 	heta)-1}{(	an 	heta-\sec 	heta)+1}=\frac{\{(	an 	heta+\sec 	heta)-1\}(	an 	heta-\sec 	heta)}{\{(	an 	heta-\sec 	heta)+1\}(	an 	heta-\sec 	heta)}$

$=\frac{\left(	an ^{2} 	heta-\sec ^{2} 	hetaight)-(	an 	heta-\sec 	heta)}{\{	an 	heta-\sec 	heta+1\}(	an 	heta-\sec 	heta)}$

$=\frac{-1-	an 	heta+\sec 	heta}{(	an 	heta-\sec 	heta+1)(	an 	heta-\sec 	heta)}$

$=\frac{-1}{	an 	heta-\sec 	heta}=\frac{1}{\sec 	heta-	an 	heta}$

which is the RHS of the identity, we are required to prove.

सर्वसमिका $\sec ^{2} \theta=1+\tan ^{2} \theta$ का प्रयोग करके सिद्ध कीजिए कि

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

Similar Questions

दिखाइए कि

$(i)$ $\tan 48^{\circ} \tan 23^{\circ} \tan 42^{\circ} \tan 67^{\circ}=1$

$(ii)$ $\cos 38^{\circ} \cos 52^{\circ}-\sin 38^{\circ} \sin 52^{\circ}=0$

$\frac{1-\tan ^{2} 45^{\circ}}{1+\tan ^{2} 45^{\circ}}=$

$(1+\tan \theta+\sec \theta)(1+\cot \theta-\operatorname{cosec} \theta)=..........$

निम्नलिखित के मान निकालिए :

$\frac{\cos 45^{\circ}}{\sec 30^{\circ}+\operatorname{cosec} 30^{\circ}}$

सर्वसमिका $\sec ^{2} \theta=1+\tan ^{2} \theta$ का प्रयोग करके सिद्ध कीजिए कि $\frac{\sin \theta-\cos \theta+1}{\sin \theta+\cos \theta-1}=\frac{1}{\sec \theta-\tan \theta}$