Trigonometry class 11 please help. handwritten correct solution will be marked brainliest
Answers
Answer:
Question: 1
Add the following:
(i) 3x and 7x
(ii) -5xy and 9xy
Solution:
We have
(i) 3x + 7x = (3 + 7) x = 10x
(ii) -5xy + 9xy = (-5 + 9)xy = 4xy
Question: 2
Simplify each of the following:
(i) 7x3y +9yx3
(ii) 12a2b + 3ba2
Solution:
Simplifying the given expressions, we have
(i) 7x3y + 9yx3 = (7 + 9)x3y = 16x3y
(ii) 12a2b + 3ba2 = (12 + 3)a2b =15a2b
Answer:
mark it as brainliest
\begin{gathered}so \: here \: \\ secA + tanA =x \\ ie \: tan \: A = x - sec \: A\end{gathered}
sohere
secA+tanA=x
ietanA=x−secA
\begin{gathered}so \: now \: using \\ 1 + tan ^{2} A = sec {}^{2} A \\ 1 + (x - sec \: A) {}^{2} = sec {}^{2} A \\ 1 + x {}^{2} + sec {}^{2} A - 2sec \: A.x = sec {}^{2} A \\ 1 + x {}^{2} - 2sec \: A.x = 0 \\ ie \: 1 + x {}^{2} = 2secA.x \\ sec \: A = 1 + x {}^{2} /2x \\ \\ \end{gathered}
sonowusing
1+tan
2
A=sec
2
A
1+(x−secA)
2
=sec
2
A
1+x
2
+sec
2
A−2secA.x=sec
2
A
1+x
2
−2secA.x=0
ie1+x
2
=2secA.x
secA=1+x
2
/2x
\begin{gathered}thus \: value \: of \: sec \: A \: (in \: terms \: of \: x) \: is \: 1 + x {}^{2} /2x \\ \end{gathered}