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Trigonometric Identity- 2

Prove that:1....

Question

Prove that:
1. $\sqrt{s e c^{2} A + c o s e c^{2} A}$ = tanA + cotA

2. $s i n^{2} A (1 + c o t^{2} A) + c o s^{2} A (1 + t a n^{2} A)$ = 2

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Solution

1. LHS = $\sqrt{1 + t a n^{2} A + 1 + c o t^{2} A}$

= $\sqrt{t a n^{2} A + c o t^{2} A + 2}$ = $\sqrt{t a n^{2} A + c o t^{2} A + t a n A . c o t A}$

= $\sqrt{(t a n A + c o t A)^{2}}$ = tanA + cotA = R.H.S

2. L.H.S = $s i n^{2} A (1 + c o t^{2} A) + c o s^{2} A (1 + t a n^{2} A)$ = $s i n^{2} A + c o s^{2} A$ + $c o s^{2} A + s i n^{2} A$ = 2

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Similar questions

\frac{{tan}^{2} A}{{tan}^{2} A + 1} + \frac{c o s e c^{2} A}{{sec}^{2} A + c o s e c^{2} A} = 1

(\frac{1 + {tan}^{2} A}{1 + {cot}^{2} A}) = {(\frac{1 - tan A}{1 - cot A})}^{2} = {tan}^{2} A

.

(\frac{1 + {tan}^{2} A}{1 + {cot}^{2} A}) = {(\frac{1 - tan A}{1 - cot A})}^{2} = {tan}^{2} A

(\frac{1 + {tan}^{2} A}{1 + {cot}^{2} A}) = {(\frac{1 - tan A}{1 - cot A})}^{2}

\sqrt{{sec}^{2} A + cos e c^{2} A} = tan A + cot A

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