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Grade:

                        

Question 2: If cos A = tan B, cos B = tan C and cos C = tan A, then sinA = Options in in the attachment Thank you

7 months ago

Answers : (2)

Arun
25185 Points
							
**cos C=tan A**
**so tan C=sqrt(1-tan ^2 A)/tan A**
**cos B=tan C=sqrt(1-tan ^2 A)/tan A**
**so tan B=sqrt((2×tan ^2 A-1)/(1-tan ^2 A))**
**But cos A=tan B**
**cos A=sqrt(2×tan ^2 A-1)/(1-tan ^2 A))**
**cos ^2 A=(2×tan ^2 A-1)/(1-tan ^2 A)**
**1/sec ^2 A=**(**2×tan ^2 A-1)/(1-tan ^2 A)**
**1/(1+tan ^2 A)=(2×tan ^2 A-1)/(1-tan ^2 A)**
**tan ^4 A +tan ^2 A -1=0**
**tan ^2 A=(-1+√5)/2**
**cot ^2 A=2/(√5 -1)**
**cosec ^2 A -1 =2/(√5 -1)**
**cosec ^2 A=(1+√5)/(√5 -1)**
**sin ^2 A=(√5–1)/(√5+1)=(√5–1)^2/(√5+1)(√5–1)**
**sin ^2 A=((√5–1)^2)/4**
**sin A=(√5–1)/2**
 
7 months ago
Vikas TU
13785 Points
							
Dear student
cos C=tan A
so tan C=sqrt(1-tan ^2 A)/tan A
cos B=tan C=sqrt(1-tan ^2 A)/tan A
so tan B=sqrt((2×tan ^2 A-1)/(1-tan ^2 A))
But cos A=tan B
cos A=sqrt(2×tan ^2 A-1)/(1-tan ^2 A))
cos ^2 A=(2×tan ^2 A-1)/(1-tan ^2 A)
1/sec ^2 A=(2×tan ^2 A-1)/(1-tan ^2 A)
1/(1+tan ^2 A)=(2×tan ^2 A-1)/(1-tan ^2 A)
tan ^4 A +tan ^2 A -1=0
tan ^2 A=(-1+√5)/2
cot ^2 A=2/(√5 -1)
cosec ^2 A -1 =2/(√5 -1)
cosec ^2 A=(1+√5)/(√5 -1)
sin ^2 A=(√5–1)/(√5+1)=(√5–1)^2/(√5+1)(√5–1)
sin ^2 A=((√5–1)^2)/4
sin A=(√5–1)/2.
7 months ago
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