Question

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grade 10 trigonometry ​

Answer 1

Step-by-step explanation:

4tan=3. tan=3

4

4 sin-cos=4tan-1

4sin+cos=4tan+1

=4×3=1.

4

4×3+1

4

=2 1

4=2

4ke sath 4cancel ho jayega

Answer 2

Given Question

$\sf \: If \: 4tan \beta = 3, \: then \: \dfrac{4sin \beta - 3cos \beta }{4sin \beta + 3cos \beta }$

$\: \: \: \: \: \: (a) \: \: 0$

$\: \: \: \: \: \: (b) \: \: \dfrac{1}{3}$

$\: \: \: \: \: \: (c) \: \: \dfrac{2}{3}$

$\: \: \: \: \: \: (d) \: \: \dfrac{3}{4}$

$\large\underline{\sf{Solution-}}$

Given that,

$\rm :\longmapsto\:4tan \beta = 3$

$\bf\implies \:tan\beta = \dfrac{3}{4}$

Now, Consider

$\rm :\longmapsto\:\dfrac{4sin\beta - 3cos\beta }{4sin\beta + 3cos\beta }$

$\rm \:  =  \: \dfrac{cos\beta \bigg[\dfrac{4sin\beta }{cos\beta } - 3\bigg] }{cos\beta \bigg[\dfrac{4sin\beta }{cos\beta } + 3 \bigg]}$

$\rm \:  =  \: \dfrac{4tan\beta - 3}{4tan\beta + 3}$

$\rm \:  =  \: \dfrac{3 - 3}{3 + 3} \: \: \: \: \: \: \: \: \: \{ \: using \: given \: 4tan\beta = 3 \}$

$\rm \:  =  \: 0$

Hence,

$\rm :\longmapsto\: \: \boxed{\bf{ \: \dfrac{4sin\beta - 3cos\beta }{4sin\beta + 3cos\beta } \: = \: 0 \: }}$

So, Option (a) is correct.

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Additional Information:-

Relationship between sides and T ratios

sin θ = Opposite Side/Hypotenuse

cos θ = Adjacent Side/Hypotenuse

tan θ = Opposite Side/Adjacent Side

sec θ = Hypotenuse/Adjacent Side

cosec θ = Hypotenuse/Opposite Side

cot θ = Adjacent Side/Opposite Side

Reciprocal Identities

cosec θ = 1/sin θ

sec θ = 1/cos θ

cot θ = 1/tan θ

sin θ = 1/cosec θ

cos θ = 1/sec θ

tan θ = 1/cot θ

Co-function Identities

sin (90°−x) = cos x

cos (90°−x) = sin x

tan (90°−x) = cot x

cot (90°−x) = tan x

sec (90°−x) = cosec x

cosec (90°−x) = sec x

Fundamental Trigonometric Identities

sin²θ + cos²θ = 1

sec²θ - tan²θ = 1

cosec²θ - cot²θ = 1