if-tan-A-cot-A-0-prove-that-sin-A-cos-A-

Question Number 60269 by Askash last updated on 19/May/19

i f

t a n A - c o t A = 0

p r o v e t h a t

s i n A + c o s A = ?

Commented by mr W last updated on 19/May/19

\sin A + \cos A = 0 o r \sqrt{2} o r - \sqrt{2}

Commented by Askash last updated on 19/May/19

h o w - \sqrt{2}

i s p o s s i b l e

Commented by mr W last updated on 19/May/19

e . g .

A = π + \frac{π}{4}

\tan A = 1

\cot A = 1

\Rightarrow \tan A - \cot A = 0

\sin A = - \frac{\sqrt{2}}{2}

\cos A = - \frac{\sqrt{2}}{2}

\Rightarrow \sin A + \cos A = - \sqrt{2}

Answered by Kunal12588 last updated on 19/May/19

\frac{s i n A}{c o s A} - \frac{c o s A}{s i n A} = 0

\Rightarrow \frac{{s i n}^{2} A - {c o s}^{2} A}{s i n A c o s A} = 0

\Rightarrow (s i n A + c o s A) (s i n A - c o s A) = 0

\Rightarrow s i n A + c o s A = 0

Commented by Askash last updated on 19/May/19

A N O T H E R W A Y

{s i n}^{2} A - {c o s}^{2} A = 0

1 - {c o s}^{2} A - {c o s}^{2} A = 0

1 - 2 {c o s}^{2} A = 0

\frac{1}{\sqrt{2}} = c o s A

c o s 45 ° = c o s A

A = 45 °

s i n 45 ° + c o s 45 °

= \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}

= \frac{2}{\sqrt{2}}

= \frac{2}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}

= \sqrt{2}

I S I T R I G H T O R W R O N G ? W H Y ?

Commented by Kunal12588 last updated on 19/May/19

{i t}^{'} s a c t u a l l y c o r r e c t

b u t i t h a s t w o a n s w e r s

1 - 2 {c o s}^{2} A = 0

\Rightarrow c o s A = \pm \frac{1}{\sqrt{2}}

y o u s o l v e d u s i n g c a s e 1 \to c o s A = \frac{1}{\sqrt{2}}

n o w I a m t a k i n g c a s e 2 \to c o s A = - \frac{1}{\sqrt{2}}

\Rightarrow c o s A = - c o s (45 °)

\Rightarrow c o s A = c o s (180 ° - 45 °) = c o s (135 °)

\Rightarrow A = 135 ° (t a k i n g o n l y p r i c i p l e v a l u e)

s i n (135 °) - c o s (135 °) = \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = 0

i n m y a n s w e r i h a d n o t t a k e n

(s i n A - c o s A) = 0

\Rightarrow s i n A + c o s A = \sqrt{2}

y e s - \sqrt{2} i s a l s o p o s s i b l e

c o s A = - c o s (45 °)

\Rightarrow c o s A = c o s (180 ° + 45 °) = c o s (225 °)

\Rightarrow A = 225 ° (i t s n o t p r i n c i p l e v a l u e)

s i n (225 °) + c o s (225 °) = - \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = - \sqrt{2}

Commented by mr W last updated on 19/May/19

w h a t a b o u t A = 225 ° ?

Commented by Kunal12588 last updated on 19/May/19

t h a n k y o u s i r

Answered by ajfour last updated on 19/May/19

\frac{p}{b} = \frac{b}{p} \Rightarrow p = b = \pm \frac{h}{\sqrt{2}} & p = - b = \pm \frac{h}{\sqrt{2}}

\sin A + \cos A = \frac{p + b}{h} = \pm \sqrt{2}, 0 .

Commented by Kunal12588 last updated on 19/May/19

g r e a t i u s e d t h i s m e t h o d t o p r o v e t h e

t r i g i d e n t i t i e s i n f i r s t y e a r b u t t e a c h e r s

a l w a y s d e d u c t m y m a r k s .

Commented by Askash last updated on 20/May/19

S a m e w i t h m e .

S o m e T e a c h e r s o n l y c h e c k t h e i r

s o l u t i o n s w h a t t h e y h a d t a u g h t .

Answered by malwaan last updated on 22/May/19

t a n A - c o t A = 0

\Rightarrow \frac{s i n A}{c o s A} - \frac{c o s A}{s i n A} = 0

\frac{{s i n}^{2} A - {c o s}^{2} A}{c o s A \times s i n A} = 0

(s i n A + c o s A) (s i n A - c o s A) = 0

1 : s i n A + c o s A = 0

2 : s i n A - c o s A = 0

\Rightarrow A = \frac{π}{4} \Rightarrow s i n A + c o s A = \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}

OR A = π + \frac{π}{4} = \frac{5 π}{4}

\Rightarrow s i n A + c o s A = - \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = - \frac{2}{\sqrt{2}} = - \sqrt{2}

∴ s i n A + c o s A = {0 \boldsymbol o r \sqrt{2} \boldsymbol o r - \sqrt{2}}

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