Question

tan theta=4/3 ho toh darsaiye root 1-sin theta/1+sin theta + root 1+sin theta 1-sin theta=10/3










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Answer 1

$\huge\pink{\boxed{\blue{\boxed{ \purple{ \boxed{{\pink{Answer}}}}}}}} \\ \large\pink{\boxed{\blue{\boxed{ \purple{ \boxed{{\pink{Your~answer↓}}}}}}}}$

Given :-

$\large\bold\red{tanθ \: = \frac{4}{3} }$

To prove :-

$\large\bold\green{ \sqrt{ \frac{1 - sinθ}{1 + sinθ} } + \sqrt{ \frac{1 + sinθ}{1 - sinθ} } = \frac{10}{3} }$

$\large\bold\green{ \sqrt{ \frac{1 - sinθ}{1 + sinθ} } + \sqrt{ \frac{1 + sinθ}{1 - sinθ} } } \\ on \: taking \: lcm \\ \large\bold\green{ = \frac{1 - sinθ \: + 1 + sinθ}{ \sqrt{(1 - sinθ)(1 + sinθ)} } } \\ \large\bold\green{ = \frac{2}{ \sqrt{1 - {sin}^{2} θ} } } \\ \large\bold\green{ = \frac{2}{ \sqrt{ {cos}^{2}θ } } } \\ \large\bold\green{ = \frac{2}{cosθ} } \\ \large\bold\red{ = 2secθ} \\ \large\bold\green{ = 2 \sqrt{1 + {tan}^{2}θ } } \\ \large\bold\green{ = 2 \sqrt{1 + {( \frac{4}{3} )}^{2} } } \\ \large\bold\green{ = 2 \sqrt{ \frac{9 + 16}{9} } } \\ \large\bold\green{ = 2 \times \frac{5}{3} } \\ \large\bold\green{ = \frac{10}{3} }$

$\large\bold\green{♧Hence, proved.♧}$

$\red{\textsf{(✪MARK BRAINLIEST✿)}} \\ \huge\fcolorbox{aqua}{aqua}{\red{FolloW ME}}$

Answer 2

Answer:10\3]

is the answer...