Question

33. Prove that (cot A + sec B)2 – (tan B - cosec A)2 = 2(cot A. sec B + tan B. cosec A)
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Answer 1

Step-by-step explanation:

∵  $(a+b)^2=(a^2+b^2+2ab)$  and  $(a-b)^2=(a^2+b^2-2ab)$

⇒$(\cot A + \sec B)^2-(\tan B - \csc A)^2$

= $(\cot^2A+sec^2B+2\cot A\sec B)-(\tan^2B+\csc^2A-2\tan B\csc A)$

= $\cot^2A-\csc^2A+\sec^2B-\tan^2B+2(\cot A\sec B+\tan B \csc A)$

∵ $\cot\theta=\frac{\cos\theta}{\sin\theta}$  , $\csc\theta=\frac{1}{\sin\theta}$ , $\sec\theta=\frac{1}{\cos\theta}$  and $\tan\theta=\frac{sin\theta}{\cos\theta}$

⇒ $\cot^2A-\csc^2A+\sec^2B-\tan^2B+2(\cot A\sec B+\tan B \csc A)$

=  $\frac{\cos^2A}{\sin^2A} -\frac{1}{\sin^2A} +\frac{1}{\cos^2B} -\frac{sin^2B}{\cos^2B} +2(\cot A\sec B+\tan B \csc A)$

= $\frac{\cos^2A-1}{\sin^2A} +\frac{1-sin^2B}{\cos^2B} +2(\cot A\sec B+\tan B \csc A)$

= $-\frac{1-\cos^2A}{\sin^2A} +\frac{1-sin^2B}{\cos^2B} +2(\cot A\sec B+\tan B \csc A)$

∵ $\sin^2\theta+\cos^2\theta=1$

⇒ $\sin^2\theta=1-\cos^2\theta$  and  $\cos^2\theta=1-\sin^2\theta$

⇒$-\frac{1-\cos^2A}{\sin^2A} +\frac{1-sin^2B}{\cos^2B} +2(\cot A\sec B+\tan B \csc A)$

= $-\frac{sin^2A}{\sin^2A} +\frac{\cos^2B}{\cos^2B} +2(\cot A\sec B+\tan B \csc A)$

= $-1+1 +2(\cot A\sec B+\tan B \csc A)$

= $2(\cot A\sec B+\tan B \csc A)$              Hence Proved