aaj mera aakhri exam h i am feeling nervous.
Answers
Answer:
tanu ji se darti hai yeh merese?
Answer:
sinθ+cosθ=a
secθ+cscθ=b
\sf\underline \red{ To\:Find}
ToFind
We have to find the value of b(a²-1)
\sf\underline \pink{ Solution }
Solution
By putting the given values
\begin{gathered}:\implies\sf\ \ b(a^2-1)\\ \\ \\ :\implies\sf\ \ sec\theta+csc\theta\big\{(sin\theta+cos\theta)^2-1)\big\}\\ \\ \\ \bullet\sf\ sec\theta=\dfrac{1}{cos\theta}\ \ ;\ csc\theta=\dfrac{1}{sin\theta}\\ \\ \\ :\implies\sf\ \dfrac{1}{cos\theta}+\dfrac{1}{sin\theta}\big\{sin^2\theta+cos^2\theta+2sin\theta cos\theta-1\big\}\\ \\ \\ \bullet\sf\ \ sin^2\theta+cos^2\theta=1\\ \\ \\ :\implies\sf\dfrac{sin\theta+cos\theta}{sin\theta\ cos\theta}\big\{\cancel{1}+2sin\theta\ cos\theta \cancel{-1}\big\}\\ \\ \\ :\implies\sf\ \dfrac{sin\theta+cos\theta}{\cancel{sin\theta cos\theta}}\times 2\cancel{sin\theta cos\theta}\\ \\ \\ :\implies\sf\ \ 2(sin\theta+cos\theta)\end{gathered}
:⟹ b(a
2
−1)
:⟹ secθ+cscθ{(sinθ+cosθ)
2
−1)}
∙ secθ=
cosθ
1
; cscθ=
sinθ
1
:⟹
cosθ
1
+
sinθ
1
{sin
2
θ+cos
2
θ+2sinθcosθ−1}
∙ sin
2
θ+cos
2
θ=1
:⟹
sinθ cosθ
sinθ+cosθ
{
1
+2sinθ cosθ
−1
}
:⟹
sinθcosθ
sinθ+cosθ
×2
sinθcosθ
:⟹ 2(sinθ+cosθ)
\underline{\bigstar{\blue{\sf\ \ b(a^2-1)= 2(sin\theta+cos\theta)}}}
★ b(a
2
−1)=2(sinθ+cosθ
Congo ki aapka aaj last exam tha mere exams to start bhi nhi huae h