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3 years ago

1) त्रिभुज LOK को क्षेत्रफल पत्ता लगाउनुहोस् । जहाँ

Mathematics

1 answer:

siniylev [52]3 years ago

7 0

Answer:

hope this helps you.keep it up..

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The quotient of 15x^4y^2-30x^2y^3+45xy/5xy is______ .When this quotient is divided by ______, the result is x^3y-2xy^2+3

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This quotient is 48divide by5

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3 years ago

X divided by 6/7 = 3 and 2/5

Rom4ik [11]

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What is an expression equivalent to -4(-6m - 7)

nekit [7.7K]

Answer: 24m+28

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3 years ago

To prove :One plus cot square theta into tan theta by sec square theta = cot theta

pickupchik [31]

we are given

$\frac{(1+cot^2(\theta))*tan(\theta)}{sec^2(\theta)} =cot(\theta)$

We will simplify left side and make it equal to right side

Left side:

$\frac{(1+cot^2(\theta))*tan(\theta)}{sec^2(\theta)}$

we can use trigonometric identity

$1+cot^2(\theta)=csc^2(\theta)$

we can replace it

$\frac{(csc^2(\theta))*tan(\theta)}{sec^2(\theta)}$

we know that

csc=1/sin and sec=1/cos

so, we can replace it

and we get

$\frac{cos^2(\theta)tan(\theta)}{sin^2(\theta)}$

now, we know that

tan =sin/cos

$\frac{cos^2(\theta)*sin(\theta)}{sin^2(\theta)*cos(\theta)}$

we can simplify it

and we get

$\frac{cos(\theta)}{sin(\theta)}$

we can also write it as

$=cot(\theta)$

Right Side:

$cot(\theta)$

we can see that

left side = right side

so,

$\frac{(1+cot^2(\theta))*tan(\theta)}{sec^2(\theta)} =cot(\theta)$ ......Answer

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3 years ago