Question

Sin 50-sin 70+sin10=o
Proved the question ​

Answer 1

$\huge\bold\red{ANSWER:-}$

We will use these two trignometric formulas to solve this.

We will use these two trignometric formulas to solve this.$1) Sin(A+B) = Sin(A)Cos(B)+Cos(A)Sin(B)$

We will use these two trignometric formulas to solve this.$1) Sin(A+B) = Sin(A)Cos(B)+Cos(A)Sin(B)$$2) Sin(A-B) = Sin(A)Cos(B)-Cos(A)Sin(B)$

We will use these two trignometric formulas to solve this.$1) Sin(A+B) = Sin(A)Cos(B)+Cos(A)Sin(B)$$2) Sin(A-B) = Sin(A)Cos(B)-Cos(A)Sin(B)$Now we write our expression as

We will use these two trignometric formulas to solve this.$1) Sin(A+B) = Sin(A)Cos(B)+Cos(A)Sin(B)$$2) Sin(A-B) = Sin(A)Cos(B)-Cos(A)Sin(B)$Now we write our expression as $Sin(60-10) - Sin(60+10) +Sin(10).[tex]</em></strong></p><p></p><p><strong><em>We will use these two trignometric formulas to solve this.[tex]1) Sin(A+B) = Sin(A)Cos(B)+Cos(A)Sin(B)$$2) Sin(A-B) = Sin(A)Cos(B)-Cos(A)Sin(B)$Now we write our expression as $Sin(60-10) - Sin(60+10) +Sin(10).[tex]On applying the above formulas we get:</em></strong></p><p></p><p><strong><em>We will use these two trignometric formulas to solve this.[tex]1) Sin(A+B) = Sin(A)Cos(B)+Cos(A)Sin(B)$$2) Sin(A-B) = Sin(A)Cos(B)-Cos(A)Sin(B)$Now we write our expression as $Sin(60-10) - Sin(60+10) +Sin(10).[tex]On applying the above formulas we get:[tex]Sin(60)Cos(10) - Cos(60)Sin(10) -{Sin(60)Cos(10) + Cos(60)Sin(10)} + Sin(10)$

We will use these two trignometric formulas to solve this.$1) Sin(A+B) = Sin(A)Cos(B)+Cos(A)Sin(B)$$2) Sin(A-B) = Sin(A)Cos(B)-Cos(A)Sin(B)$Now we write our expression as $Sin(60-10) - Sin(60+10) +Sin(10).[tex]On applying the above formulas we get:[tex]Sin(60)Cos(10) - Cos(60)Sin(10) -{Sin(60)Cos(10) + Cos(60)Sin(10)} + Sin(10)$On simplifying we get:

We will use these two trignometric formulas to solve this.$1) Sin(A+B) = Sin(A)Cos(B)+Cos(A)Sin(B)$$2) Sin(A-B) = Sin(A)Cos(B)-Cos(A)Sin(B)$Now we write our expression as $Sin(60-10) - Sin(60+10) +Sin(10).[tex]On applying the above formulas we get:[tex]Sin(60)Cos(10) - Cos(60)Sin(10) -{Sin(60)Cos(10) + Cos(60)Sin(10)} + Sin(10)$On simplifying we get:$-2*Cos(60)Sin(10) + Sin(10)$

We will use these two trignometric formulas to solve this.$1) Sin(A+B) = Sin(A)Cos(B)+Cos(A)Sin(B)$$2) Sin(A-B) = Sin(A)Cos(B)-Cos(A)Sin(B)$Now we write our expression as $Sin(60-10) - Sin(60+10) +Sin(10).[tex]On applying the above formulas we get:[tex]Sin(60)Cos(10) - Cos(60)Sin(10) -{Sin(60)Cos(10) + Cos(60)Sin(10)} + Sin(10)$On simplifying we get:$-2*Cos(60)Sin(10) + Sin(10)$$Using Cos(60) = 1/2 in above expression we get:$

We will use these two trignometric formulas to solve this.$1) Sin(A+B) = Sin(A)Cos(B)+Cos(A)Sin(B)$$2) Sin(A-B) = Sin(A)Cos(B)-Cos(A)Sin(B)$Now we write our expression as $Sin(60-10) - Sin(60+10) +Sin(10).[tex]On applying the above formulas we get:[tex]Sin(60)Cos(10) - Cos(60)Sin(10) -{Sin(60)Cos(10) + Cos(60)Sin(10)} + Sin(10)$On simplifying we get:$-2*Cos(60)Sin(10) + Sin(10)$$Using Cos(60) = 1/2 in above expression we get:$$-2*(1/2)*Sin(10) + Sin(10)$

We will use these two trignometric formulas to solve this.$1) Sin(A+B) = Sin(A)Cos(B)+Cos(A)Sin(B)$$2) Sin(A-B) = Sin(A)Cos(B)-Cos(A)Sin(B)$Now we write our expression as $Sin(60-10) - Sin(60+10) +Sin(10).[tex]On applying the above formulas we get:[tex]Sin(60)Cos(10) - Cos(60)Sin(10) -{Sin(60)Cos(10) + Cos(60)Sin(10)} + Sin(10)$On simplifying we get:$-2*Cos(60)Sin(10) + Sin(10)$$Using Cos(60) = 1/2 in above expression we get:$$-2*(1/2)*Sin(10) + Sin(10)$$= 0$

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