Question

pls answer, I am going to mark correct answer as brainliest​

Answer 1

Check the attachment for answer and mark me brainliest

Answer 2

Question:

Prove that (sinθ + secθ)² + (cosθ + cosecθ)² = (1 + secθcosecθ)²

Step-by-step explanation:

L.H.S. = (sinθ + secθ)² + (cosθ + cosecθ)²

___________________________

On further solving, we get

___________________________

${\boxed{\sf{\red{Identity \ : \ sec \theta = {\dfrac{1}{cos \theta}} }}}}$

${\boxed{\sf{\red{Identity \ : \ cosec \theta = {\dfrac{1}{sin \theta}}}}}}$

___________________________

$\Rightarrow{\sf{ \left( sin \theta + {\dfrac{1}{cos \theta}} \right) ^2 + \left( cos \theta + {\dfrac{1}{sin \theta}} \right) ^2 }}$

__________________________

$\Rightarrow{\sf{ \left( {\dfrac{sin \theta cos \theta + 1}{cos \theta }} \right) ^2 + \left( {\dfrac{sin \theta cos \theta + 1}{ sin \theta }} \right) ^2 }}$

__________________________

$\Rightarrow{\sf{ {\dfrac{ (sin \theta cos \theta + 1)^2}{ (cos \theta)^2}} + {\dfrac{ (sin \theta cos \theta + 1)^2}{(sin \theta)^2}} }}$

__________________________

$\Rightarrow{\sf{ {\dfrac{ (sin \theta cos \theta + 1)^2}{ cos^2 \theta}} + {\dfrac{ (sin \theta cos \theta + 1)^2}{sin^2 \theta}} }}$

__________________________

$\Rightarrow{\sf{ (sin \theta cos \theta + 1)^2 \left( {\dfrac{1}{cos^2 \theta}} + {\dfrac{1}{sin^2 \theta}} \right) }}$

___________________________

$\Rightarrow{\sf{ (sin \theta cos \theta + 1)^2 \left( {\dfrac{sin^2 \theta + cos^2 \theta}{cos^2 \theta sin^2 \theta}} \right) }}$

____________________________

${\boxed{\sf{\red{Identity \ : \ (a + b)^2 = a^2 + 2ab + b^2}}}}$

${\sf{\red{Here, \ a = sin \theta cos \theta , \ b = 1}}}$

____________________________

${\boxed{\sf{\red{Identity \ : \ sin^2 \theta + cos^2 \theta = 1}}}}$

____________________________

$\Rightarrow{\sf{ [ (sin \theta cos \theta )^2 + 2(sin \theta cos \theta)(1) + (1)^2 ] \left( {\dfrac{1}{cos^2 \theta sin^2 \theta }} \right) }}$

_____________________________

$\Rightarrow{\sf{ (sin^2 \theta cos^2 \theta + 2sin \theta cos \theta + 1) \left( {\dfrac{1}{cos^2 \theta sin^2 \theta }} \right) }}$

______________________________

$\Rightarrow{\sf{sin^2 \theta cos^2 \theta \left( {\dfrac{1}{cos^2 \theta sin^2 \theta}} \right) + 2sin \theta cos \theta \left( {\dfrac{1}{cos^2 \theta sin^2 \theta }} \right) + 1 \left( {\dfrac{1}{cos^2 \theta sin^2 \theta}} \right) }}$

_______________________________

$\Rightarrow{\sf{{\cancel{sin^2 \theta cos^2 \theta}} \left( {\dfrac{1}{{\cancel{cos^2 \theta sin^2 \theta}}}} \right) + 2sin \theta cos \theta \left( {\dfrac{1}{cos^2 \theta sin^2 \theta }} \right) + 1 \left( {\dfrac{1}{cos^2 \theta sin^2 \theta}} \right) }}$

____________________________

$\Rightarrow{\sf{ 1 + \left( {\dfrac{2sin \theta cos \theta}{cos^2 \theta sin^2 \theta}} \right) + \left( {\dfrac{1}{cos^2 \theta sin^2 \theta}} \right) }}$

_____________________________

$\Rightarrow{\sf{ 1 + {\dfrac{2}{cos \theta sin \theta}} + {\dfrac{1}{cos^2 \theta sin^2 \theta}} }}$

_____________________________

${\boxed{\sf{\red{Identity \ : \ sec \theta = {\dfrac{1}{cos \theta}} }}}}$

${\boxed{\sf{\red{Identity \ : \ cosec \theta = {\dfrac{1}{sin \theta}} }}}}$

______________________________

$\Rightarrow{\sf{1 + 2sec \theta cosec \theta + sec^2 \theta cosec^2 \theta}}$

______________________________

${\boxed{\sf{\red{Identity \ : \ a^2 + 2ab + b^2 = (a + b)^2}}}}$

${\sf{\red{Here, \ a = 1, \ b = sec \theta cosec \theta}}}$

_____________________________

$\Rightarrow{\sf{(1 + sec \theta cosec \theta)^2}}$

____________________________

= R.H.S.

Hence, proved !!