Question

please solve the trigonometrical question
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Answer 1

             

$\sec^2\theta-\tan^2\theta=1 \\ \\ \therefore\ \tan^2\theta-\sec^2\theta=-1 \\ \\ \\ \sec\theta \times \cos\theta \ = \ \frac{1}{\cos\theta} \times \cos\theta \ = \ \frac{\cos\theta}{\cos\theta} \ = \ 1 \\ \\ OR \\ \\ \sec\theta \times \cos\theta \ = \ \sec\theta \times \frac{1}{\sec\theta} \ = \ \frac{\sec\theta}{\sec\theta} \ = \ 1 \\ \\ \\ \\ \frac{\sin\theta}{\cos\theta}=\tan\theta \\ \\ \therefore\ \sin\theta=\tan\theta \times \cos\theta$

         

       

$\boxed{LHS} \\ \\ \\ \boxed{\frac{\tan\theta-\sec\theta+1}{\tan\theta+\sec\theta-1}} \\ \\ \\ \boxed{\frac{(\tan\theta-\sec\theta+1)(\tan\theta+\sec\theta)}{(\tan\theta+\sec\theta-1)(\tan\theta+\sec\theta)}} \\ \\ \\ \boxed{\frac{(\tan\theta-\sec\theta)(\tan\theta+\sec\theta)+1(\tan\theta+\sec\theta)}{(\tan\theta+\sec\theta-1)(\tan\theta+\sec\theta)}} \\ \\ \\ \boxed{\frac{\tan^2\theta-\sec^2\theta+\tan\theta+\sec\theta}{(\tan\theta+\sec\theta-1)(\tan\theta+\sec\theta)}}$

$\boxed{\frac{-1+\tan\theta+\sec\theta}{(\tan\theta+\sec\theta-1)(\tan\theta+\sec\theta)}} \\ \\ \\ \boxed{\frac{\tan\theta+\sec\theta-1}{(\tan\theta+\sec\theta-1)(\tan\theta+\sec\theta)}} \\ \\ \\ \boxed{\frac{1}{\tan\theta+\sec\theta}} \\ \\ \\ \boxed{\frac{1}{\sec\theta+\tan\theta}} \\ \\ \\ \boxed{\frac{1 \times \cos\theta}{(\sec\theta+\tan\theta)\cos\theta}}$

$\boxed{\frac{\cos\theta}{\sec\theta \times \cos\theta +\tan\theta \times \cos\theta}} \\ \\ \\ \boxed{\frac{\cos\theta}{1+\sin\theta}} \\ \\ \\ \boxed{RHS}$

     

$Hence proved! \\ \\ \\ Please mark it as the brainliest. \\ \\ \\ Thank you. :-)$