Question

Maths Trigonometry 【 iii 】

Answer 1

Answer:

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Answer 2

$\huge{\mathtt\: SOLUTION \mapsto }\ \\ \\ \bf\ sec \theta = x + \frac{1}{4x} \\ \\ \sf\ \implies: {sec}^{2} \theta = {(x + \frac{1}{4x} })^{2} \\ \\ \sf\ \implies: {sec}^{2}\theta = {x}^{2} + 2 \times x \times \frac{1}{4x} + { (\frac{1}{4x} })^{2} \\ \\ \sf\ \implies:{sec}^{2}\theta = {x}^{2} + \frac{1}{2} + \frac{1}{ {16x}^{2} } \\ \\ \tt{\red{ \bigstar \underline{1 + {tan}^{2} \theta ={sec}^{2}\theta }}} \\ \tt{\purple {\bigstar{ \underline{{sec}^{2} - 1= \boxed{\bf{tan}^{2}\theta} }}}} \\ \\ \sf\ \implies: {tan}^{2} \theta = {x}^{2} + \frac{1}{2} + \frac{1}{ {16x}^{2}} - 1\\ \\ \sf\ \implies:{tan}^{2} \theta = {x}^{2} + \frac{1}{ {16x}^{2} } + \frac{1}{2} - 1 \\ \\ \sf\ \implies:{tan}^{2} \theta = {(x - \frac{1}{4x} )}^{2} \\ \\ \tt\ \implies: tan \theta = ± \sqrt{ {(x - \frac{1}{4x} })^{2} } \\ \\ \boxed{\tt\ \implies: tan \theta = ± \: x - \frac{1}{4x} } \\ \\ \bf\ \therefore\: tan \theta + sec \theta = x \: \: \cancel{- \frac{1}{4x}} \: + x \: \: \cancel{+ \frac{1}{4x}} \\ \\ \boxed{\bf\ \: tan \theta + sec \theta = \red2 \pink x \: \ddot \smile} \\ \\ \bf\ \therefore\: tan \theta + sec \theta = \cancel{- x} \: \: + \frac{1}{4x} \: \: \cancel{+ x} + \frac{1}{4x} \\ \\\bf\ \implies \: tan \theta + sec \theta = \frac{1 + 1}{4x} = \frac{2}{4x} \\ \\\boxed{\bf\ \: tan \theta + sec \theta = \frac{ \blue1}{ \orange2 \purple{x} } \ddot \smile }$