Let x = cot β + 5 tan β2⇒x = cot2β + 25 tan2β + 2 × cot β × 5 tan β⇒x = cot2β + 25 tan2β +10 cot β tan β⇒x = cot2β + 25 tan2β + 10 × 1tan β× tan β⇒x = cot2β + 25 tan2β + 10⇒x = cot2β + 25 tan2β + 10 - 10 + 10 Adding and subtracting 10⇒x = cot2β + 25 tan2β - 10 + 10 + 10⇒x = cot2β + 25 tan2β - 10 +20⇒x = cot β2 + 5 tan β2 - 2 × cot β × 5 tan β + 20 as, cot β tan β = 1⇒x = cot β - 5 tan β2 + 20⇒cot β + 5 tan β2 = cot β - 5 tan β2 + 20

pls help fast!!!!

Let x = {(\cot β + 5 \tan β)}^{2} \Rightarrow x = \cot^{2} β + 25 \tan^{2} β + 2 \times \cot β \times 5 \tan β \Rightarrow x = \cot^{2} β + 25 \tan^{2} β + 10 \cot β . \tan β \Rightarrow x = \cot^{2} β + 25 \tan^{2} β + 10 \times \frac{1}{\tan β} \times \tan β \Rightarrow x = \cot^{2} β + 25 \tan^{2} β + 10 \Rightarrow x = \cot^{2} β + 25 \tan^{2} β + 10 - 10 + 10 (Adding and subtracting 10) \Rightarrow x = \cot^{2} β + 25 \tan^{2} β - 10 + 10 + 10 \Rightarrow x = [\cot^{2} β + 25 \tan^{2} β - 10] + 20 \Rightarrow x = [{(\cot β)}^{2} + {(5 \tan β)}^{2} - 2 \times \cot β \times 5 \tan β] + 20 [as, \cot β . \tan β = 1] \Rightarrow x = {(\cot β - 5 \tan β)}^{2} + 20 \Rightarrow {(\cot β + 5 \tan β)}^{2} = {(\cot β - 5 \tan β)}^{2} + 20

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​pls help fast!!!!

pls help fast!!!!