Question

Please solve the question shown in the picture


Guys please solve it!!!!

मैंने यह सेक्ंद टाइम भेजा है।

क्या किसी को यह सोल्व करना नहीं आता ?
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Answer 1

Answer:

tan x - sec x [ 3 rd quadrant ]

- ( sec x + tan x ) [ 2 nd quadrant ]

Step-by-step explanation:

π/2 < x < 3 π/2

This means that x is present in the 2 nd and the 3 rd quadrants .

$\sqrt{\frac{1-sinx}{1+sinx}}\\\\\implies \sqrt{\frac{1-sinx}{1+sinx}\times \frac{1-sinx}{1-sinx}}\\\\\implies \sqrt{\frac{(1-sinx)^2}{1-sin^2x}}\\\\\implies \sqrt{\frac{(1-sinx)^2}{cos^2x}}\\\\\implies \frac{1-sinx}{cosx}$

ALL - SIN - TAN - COS rule

Sin is positive in 2 nd quadrant and negtive .

( 1 - sin x ) / cos x

= 1 /  cos x - sin x / cos x

= sec x - tan x

Note that sec x is negative in both 2 nd and in 3 rd .

tan x is positive in 3 rd but negative in 2 nd .

Hence there will be 2 solutions given in ANSWERS .

Answer 2

ANSWER:------------

π/2 < x < 3 π/2

1−sinx1+sinx⟹1−sinx1+sinx×1−sinx

1−sinx⟹(1−sinx)21−sin2x⟹(1−sinx)2c

os2x

⟹1−sinxcosx\begin{lgathered}\sqrt{\frac{1-sinx}

{1+sinx}}\\\\\implies \sqrt{\frac{1-sinx}

{1+sinx}\times \frac{1-sinx}{1-sinx}}\\\\\implies \sqrt{\frac{

(1-sinx)^2}{1-sin^2x}}\\\\\implies \sqrt{\frac{(1-

sinx)^2}{cos^2x}}\\\\\implies \frac{1-

sinx}{cosx}\end{lgathered}1+sinx1−

sinx⟹1+sinx1−sinx×1−sinx1−sinx

⟹1−sin2x(1−sinx)2⟹cos2x(1−sinx)2⟹cosx1−sinx

hope it helps:-----

T!—!ANKS!!!