Question

Sec 3A=cosec(A-10)where 3A is an acute angle find the value of A

Answer 1

${\large{\underline{\boxed{\tt{\color{lime}{Required Answer:}}}}}}$

To Solve:

• Sec 3A=cosec(A-10)where 3A is an acute angle find the value of A

Solⁿ:

Sec 3A=cosec(A-10)

(∵ secA = cosec (90-A))

cosec( 90 - 3A) = cosec (A - 10)

90 - 3A = A - 10

90 + 10 = 4A

100 = 4A

100/4 = A

25 = A

The Value of A is 25

HOPE IT HELPS

MARK BRAINLIEST PLS :)

Answer 2

$\sf{\green{\underline{\underline{\orange{QUESTION}}}}}$

• (1+cota+tana)(sina-cosa) = (sec^3a-cosec^3a)/(sec^2a*cosec^2a)

$\bold{\boxed{\small{\boxed{\orange{\small{\boxed{\small{\red{\bold{\:Answer}}}}}}}}}}$

Solving LHS ,

$\begin{gathered}\sf(1+cota+tana)(sina-cosa) \\ \\\sf (1 + \frac{cosa}{sina} + \sf\frac{sina}{cosa}) \: (sina-cosa) \\\\ \sf(sina + cosa + \frac{sin ^{2}a }{cosa} ) - (cosa + \frac{cos^{2}a }{sina} + sina) \\\ \\ \sf\frac{sin ^{2}a }{cosa} \: - \frac{cos^{2}a }{sina} \: \\ \sf\\ \sf\frac{sin^{2}a }{1} \times \frac{1}{cosa} - \frac{cos^{2}a }{1} \times \frac{1}{sina} \\ \\ \sf\frac{seca}{cosec^{2}a } - \: \frac{coseca}{sec^{2}a } \\ \\ \sf\frac{(sec^{3}a - cosec^{3}a)}{cosec^{2}a \times sec^{2} a } \\ \\ \end{gathered}$