Question

(secA-tanA)^2(1+sinA)=1-sinA​

Answer 1

Answer:

at what class you are pursuing

Step-by-step explanation:

in 10th or 11th

Answer 2

S O L U T I O N :

$\underline{\bf{Given\::}}$

(sec A - tan A)² (1 + sin A) = 1 - sin A

$\underline{\bf{Explanation\::}}$

Using some identities a/q;

• sec Ф = 1/cos Ф
• tan Ф = sin Ф/cos Ф

Taking L.H.S :

$\mapsto\tt{(sec A-tanA)^{2} (1+sinA)}$

$\mapsto\tt{(secA)^{2} + (tanA)^{2} -2\times secA .tanA \times (1+sinA)\:\:\underbrace{\sf{Using\:formula \:(a+b)^{2} }}}$

$\mapsto\tt{\dfrac{1}{cos^{2} A} + \dfrac{sin^{2}A}{cos^{2}A} - 2\times \dfrac{1}{cosA} \times \dfrac{sinA}{cosA} \times (1+sinA)}$

$\mapsto\tt{\dfrac{1+sin^{2}A}{cos^{2}A} -\dfrac{2sinA}{cos^{2}A} \times (1+sinA)}$

$\mapsto\tt{\dfrac{1+sin^{2}A - 2sinA}{cos^{2}A} \times (1+sinA)}$

$\mapsto\tt{\dfrac{(1-sinA)^{2} \times 1+sinA}{1-sin^{2}A} \:\:\underbrace{\sf{cos^{2}\theta = 1-sin^{2}\theta }}}$

$\mapsto\tt{\dfrac{(1-sinA) \times (1-sinA) \times \cancel{1+sinA}}{\cancel{(1+sinA)}(1-sinA)} }$

$\mapsto\tt{\dfrac{(1-sinA)\cancel{(1-sinA)}}{\cancel{(1-sinA)}} }$

$\mapsto\bf{(1-sinA)}$

Hence,

L.H.S = R.H.S

Proved .