Question

prove that sin theta + cosec theta whole square + cos theta + sec theta whole square equals to 7 + tan square theta + cot square theta

Answer 1

Answer:

Concept:

Trigonometric Identities: There are three primary trigonometric ratios sin, cos, and tan. The three other trigonometric ratios sec, cosec, and cot in trigonometry are the reciprocals of sin, cos, and tan respectively.

Step-by-step explanation:

Given:

$(\sin \theta+\csc \theta)^{2}+(\cos \theta+\sec \theta)^{2}=7+\tan ^{2} \theta+\cot ^{2} \theta$

Find:

Prove  that   $(\sin \theta+\csc \theta)^{2}+(\cos \theta+\sec \theta)^{2}=7+\tan ^{2} \theta+\cot ^{2} \theta$

Solution:

$\text { L.H.S. }= (\sin \theta+\csc \theta)^{2}+(\cos \theta+\sec \theta)^{2}$

           $= \sin ^{2} \theta+\csc^{2} \theta+2 \sin \theta \cdot \csc \theta+\cos ^{2} \theta+\sec ^{2} \theta+2 \cos \theta \cdot \sec \theta$

            $=\left(\sin ^{2} \theta+\cos ^{2} \theta\right)+1+\cot ^{2} \theta+2+1+\tan ^{2} \theta+2\\\\=1+6+\tan ^{2} \theta+\cot ^{2} \theta$

            $= 7+\tan ^{2} \theta+\cot ^{2} \theta$

            = R.H.S

Hence Proved.

$(\sin \theta+\csc \theta)^{2}+(\cos \theta+\sec \theta)^{2}=7+\tan ^{2} \theta+\cot ^{2} \theta$

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Answer 2

Step-by-step explanation:

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