If θ is an acute angle and 5 𝑐𝑜𝑠𝑒𝑐 θ = 7, then evaluate 𝑠𝑖𝑛θ + 𝑐𝑜𝑠2θ − 1.

# If θ is an acute angle and 5 𝑐𝑜𝑠𝑒𝑐 θ = 7, then evaluate 𝑠𝑖𝑛θ + 𝑐𝑜𝑠2θ − 1.

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### Solution:

We have been given that 5 𝑐𝑜𝑠𝑒𝑐 θ = 7, where θ is an acute angle and we have to find the value of 𝑠𝑖𝑛θ + 𝑐𝑜𝑠2θ − 1.

$\begin{array}{r}5\mathrm{cosec}\theta =7\\ ⇒\mathrm{cosec}\theta =\frac{7}{5}\end{array}$

We know that 𝑐𝑜𝑠𝑒𝑐 θ = 1 / 𝑠𝑖𝑛θ

Therefore, 𝑠𝑖𝑛 θ = 5/7

We also know that ${\mathrm{sin}}^{2}\theta +{\mathrm{cos}}^{2}\theta =1$

$\begin{array}{l}⇒{\mathrm{cos}}^{2}\theta =11-{\mathrm{sin}}^{2}\theta \\ ⇒{\mathrm{cos}}^{2}\theta =1-{\left(\frac{5}{7}\right)}^{2}\\ {\mathrm{cos}}^{2}\theta =\frac{24}{49}\end{array}$

Substituting these values in the expression 𝑠𝑖𝑛θ + 𝑐𝑜𝑠2 θ, we get

$\begin{array}{ll}\mathrm{sin}& \theta +{\mathrm{cos}}^{2}\theta -1=\frac{5}{7}+\frac{24}{49}-1=\frac{10}{49}\end{array}$

Therefore the value of the given expression is $\frac{10}{49}$.  +91

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