Question

Sin 45°×Cos 65°+Sin 135°×Cos115° = ?
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Answer 1

Answer:

sin45° ×cos(90°-45°) + sin(180°-45°) ×cos(90°+45° )

sin45° ×sin45° + cos45° × sin45° ×(-sin45° )

sin²45° - cos²45°

(1/√2)² - (1/√2)²

1/2 - 1/2

1 - 1/2 = 0/2 = 0

Step-by-step explanation:

hope it will help you

Answer 2

$\huge\purple{ANSWER:-}$

$\huge\red{G}\pink{i}\orange{v}\green{e}\blue{n:-}$

$\small\red {=> sin45°cos65° + sin135°cos115°=0}$

$\small\pink{L.H.S. \: \: sin45°cos65° + sin135°cos115°}$

$\small\orange{We \: know \: that}$

$\small\green{ sin (130°-θ)= sinθ ; </p><p>//cos(180 °- θ)= - cosθ }$

$\small\blue{Now: sin45°cos65° + </p><p> sin(180° - 45°) cos (180°-65°)}$

$=> sin45°.cos65° +sin45°(-cos65°)$

$\small\red{=> sin45° cos65° - sin45° cos65°}$

$\small\purple{ = > 0 = R.H.S.}$

$\small \pink{---Hence Proved---}$

$\huge\red{@MNF}$